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Chap 05 SM Page 171 Thursday, October 12, 2000 10:22 AM

Differential calculus

5 VCE coverage Area of study Units 3 & 4 • Calculus

In this chapter 5A The derivative of tan kx 5B Second derivatives 5C Analysing the behaviour of functions using the second derivative 5D Derivatives of inverse circular functions 5E Antidifferentiation involving inverse circular functions

Chap 05 SM Page 172 Thursday, October 12, 2000 10:22 AM

172

Specialist Mathematics

Introduction In this chapter we extend the functions which can be differentiated to include f(x) = tan kx and the inverse circular functions Sin−1x, Cos−1x and Tan−1x. It is assumed that the standard results for the differential calculus are familiar to the student as are the product rule, quotient rule and chain rule. These results are listed in the table below.

f′(x)

f(x) ax n

anx n – 1

loge kx

1 --x-

e kx

ke kx

sin kx

k cos kx

cos kx

–k sin kx

u(x) × v(x)

u′(x) × v(x) + u(x) × v′(x) (product rule)

u( x) ----------v( x)

v′ ( x ) - (quotient rule) u′(x) × v(x) - u(x) × ----------v2

g[h(x)]

h′(x) × g′[h(x)] (chain rule)

The derivative of tan kx sin kx Since tan kx = --------------- , its derivative can be found using the quotient rule: cos kx dv du v ------ – u -----d  u dx dx ------ --- = -----------------------dx  v v2 If f(x) = tan kx, sin kx = --------------cos kx where k is a constant, then using the quotient rule with u = sin kx and v = cos kx: k cos kx ( cos kx ) – sin kx ( – k sin kx ) f ′(x) = ----------------------------------------------------------------------------------------cos 2 kx k cos 2 kx + k sin 2 kx = ----------------------------------------------cos 2 kx k ( cos 2 kx + sin 2 kx ) = --------------------------------------------cos 2 kx k(1) = --------------cos 2 kx

(by factorising the numerator)

(since cos2kx + sin2kx = 1)

= k sec2kx If f(x) = tan kx, then f ′(x) = k sec2 kx.

Chap 05 SM Page 173 Thursday, October 12, 2000 10:22 AM

Chapter 5

Differential calculus

173

WORKED Example 1

Differentiate the following expressions with respect to x. 4x a y = tan 6x b y = 2 tan -----3 THINK WRITE a y = tan 6x a Differentiate by rule: y = tan kx, dy dy ------ = 6 sec26x then ------ = ksec2kx where k = 6. dx dx 4x 4x b 1 Differentiate tan ------ by rule where b y = 2 tan -----3 3 k = 4--3- . 2

dy 4 4x ------ = 2  --- sec 2 ------ 3 dx 3 dy 8 4x ------ = --- sec 2 -----dx 3 3

Multiply the derivative by 2.

WORKED Example 2

If f(x) = tan (x2 + 5x), find f ′(x). THINK 1

2 3 4 5

6

WRITE

Let u = x + 5x to apply the chain rule. 2

du Find ------ . dx Express f(x) in terms of u. dy Find ------ . du dy Find ------ using the chain rule. dx Replace u with the expression x2 + 5x.

f(x) = tan(x2 + 5x) Let u = x2 + 5x. du ------ = 2x + 5 dx So f(x) = y = tan u. dy ------ = sec2u du dy dy du ------ = ------ × -----dx du dx = (2x + 5) sec2u f ′(x) = (2x + 5) sec2(x2 + 5x)

If y = tan [f(x)] dy then ------ = f ′(x) sec2 [f(x)]. dx

WORKED Example 3 dy Find ------ if y = loge [tan (−3x)]. dx THINK 1 Let u = tan (−3x) and consequently apply the chain rule. du 2 Find ------ . dx

WRITE y = loge [tan(−3x)] Let u = tan(−3x). du ------ = −3 sec2(−3x) dx

Continued over page

Chap 05 SM Page 174 Thursday, October 12, 2000 10:22 AM

174

Specialist Mathematics

THINK 3

Express y in terms of u.

4

dy Find ------ . du

5

dy Find ------ using the chain rule. dx

6

Replace u with the expression tan (−3x).

WRITE y = logeu dy 1 ------ = --du u dy dy du So ------ = ------ × -----dx du dx 1 = --- × −3 sec2 (–3x) u 2

dy Express ------ in terms of sin (−3x) and dx cos (−3x) only in order to simplify the rational expression.

7

– 3sec ( – 3x ) = ------------------------------tan ( – 3x ) –3 sin ( – 3x ) - ÷ ----------------------= -----------------------cos ( – 3x ) cos 2 ( – 3x )

–3 cos ( – 3x ) - × ----------------------= -----------------------2 sin ( – 3x ) cos ( – 3x )

8

Express the division of rational expressions as a multiplication.

9

Cancel out the factor of cos (−3x).

–3 = --------------------------------------------sin ( – 3x ) cos ( – 3x )

10

Use the double angle formula to simplify sin (−3x) cos (−3x).

–3 = ----------------------1 --- sin ( – 6x ) 2

11

State the answer in simplest form.

dy 6 and hence ------ = – ---------------------- or −6 cosec(−6x). dx sin ( – 6x )

WORKED Example 4

π Find the equation of the tangent to the curve y = 3x + cos 2x + tan x where x = --- . 4 THINK WRITE 1

To find the equation of a tangent line to a curve at a point, the coordinate of the point is needed as is the gradient of the curve at that point. Find y when x = --π4 to establish the coordinate.

y = 3x + cos 2x + tan x 3π π π If x = --- , y = ------ + cos --4 4 2 3π = ------ + 0 + 1 4 3π = 1 + -----4 π The coordinate is --- , 1 + 4 dy ------ = 3 − 2 sin 2x + sec2x dx

(

2

dy Find ------ to establish the gradient dx function.

π + tan --4

3π -----4

)

Chap 05 SM Page 175 Thursday, October 12, 2000 10:22 AM

Chapter 5

THINK 3

Differential calculus

175

WRITE

dy π Evaluate ------ when x = --- to find the 4 dx π gradient of the tangent at x = --- . 4

π dy π π If x = --- , ------ = 3 − 2 sin --- + sec2 --4 dx 2 4 1 = 3 − 2 + --------------π cos 2 --4 1 = 1 + ---------

( 1--2- )

4

5

π 3π Substitute y1 = 1 + ------ , x1 = --- and 4 4 m = 3 into the equation for a straight line: y − y1 = m(x − x1) where m is the gradient and (x1, y1) is a point on the line.

=3 so the equation of the tangent is: 3π π y − 1 + ------ = 3 x − --4 4

(

Simplify the equation and make y the subject.

) (

)

3π 3π y = 3x − ------ + 1 + -----4 4 y = 3x + 1

remember remember dy 1. If y = tan kx then ------ = k sec2 kx. dx dy 2. If y = tan f(x) then ------ = f ′(x) sec2 [f (x)]. dx

5A WORKED

1

1 Differentiate each of the following with respect to x. a y = tan 4x b y = tan 5x d y = 4 tan 2x e y = tan(−3x) g y = 3 tan(−4x) h y = −2 tan 3x

c f i

y = 3 tan 7x y = tan(−12x) y = −5 tan(−2x)

x k y = tan --2

l

3x y = tan -----5

2x m y = tan -----7

3x n y = 8 tan -----4

4x o y = 6 tan -----9

5x p y = −3 tan -----6

8x q y = 5 tan -----5

r

j

x y = tan --5

7x y = −4 tan -----2

Math

cad

Example

The derivative of tan kx

Differentiator

Chap 05 SM Page 176 Thursday, October 12, 2000 10:22 AM

176 WORKED WORKED

Example Example

2

Specialist Mathematics

2 For each of the following find f ′(x) if f(x) equals: a tan(x2 + 3x) b tan(x + 2) d tan(2x2 − 3x) e tan(3x + 2) g tan(7 − 4x) h tan(1 − 5x2) 3 multiple choice If y = etan 2x, then: dy a ------ can be found by using: dx A a direct rule B D the quotient rule E dy b ------ is equal to: dx B A 2 sec22xetan 2x D 2 tan2xesec 2x E

WORKED

Example

3

c f

tan(5x − 4) tan 8x

the product rule graphical methods

C the chain rule

2 sec2xetan 2x e x sec22x

C 4 sec2xetan 2x

dy 4 For each of the following find ------ if y equals: dx a loge(tan 6x) b etan 3x c tan(e3x) d sin(tan 5x) e cos(tan 2x) f 1--2- (tan 5x) 6x g sin 4x − tan 3x h loge2x2 + 4 tan -----7 j tan42x i tan2x 5 For each of the following find f ′(x) if f (x) equals 4x a x2 tan 3x b cos 2x tan -----5 2

d e4xtan(−3x)

e e4x tan 8x

4x 3 g -------------tan 2x

h

c

(5x3 − 6x) tan 4x

f

tan x ---------2x 2

sin 4x -------------tan 4x

6 If f: [0, π ] → R, f (x) = tan x, find the coordinates of the points on the graph where the gradient is equal to: a 1 b 4--3c 4 d 10 x 7 If f: [− π , π ] → R, f(x) = 4 tan --- , find the coordinates of the points on the graph 2 where the gradient is equal to: a 4 b 2 c 0 π WORKED 8 Find the equation of the tangent to the curve with equation y = 3 tan 2x at x = --- . Example 6 4 9 If f(x) = x tan x find the equation of i the tangent and ii the normal at the point on the curve where x = --π- . 4 10 Show that there are no stationary points for the graph of y = tan x for all values of x. 11 Explain why the gradient of tan x is always positive. π π 12 If f: − --- , --- → R, f (x) = tan x − x: 2 2 a find any stationary points and state their nature b if the domain is changed to R, show that the gradient can never be negative.

[

]

Chap 05 SM Page 177 Thursday, October 12, 2000 10:22 AM

Chapter 5

Differential calculus

177

Second derivatives Suppose that y = f (x). dy Then the derivative, also known as the first derivative, is written as f ′(x) or ------ . dx 2 d y By differentiating a second time, the second derivative, f ′′(x) or -------2- is obtained. dx 2 d y The process of obtaining f ′′(x) or -------2- from y = f (x) is also called double differendx tiation. The second derivative is commonly called ‘f double dash’ or ‘d squared y, dx squared’.

WORKED Example 5

d2 y Find --------2- if y = dx THINK 1

Express

x + 2x4 − x−1. WRITE 1 ---

x as x 2 .

x + 2x4 − x−1

y=

1 ---

y = x 2 + 2x4 − x−1 2

dy Find the first derivative ------ . dx

3

dy d2 y Differentiate ------ to obtain -------2- . dx dx

dy ------ = dx

1 --2

x

1 2

– ---

+ 8x3 + x−2

3 – --d2 y -------2- = − 1--4- x 2 + 24x2 − 2x−3 dx 2 1 or − --------3 + 24x2 − ----3--x 4x 2

WORKED Example 6

Find i f ′(x) and ii f ′′(x) if f (x) is equal to: sin x b ----------- . a ecos 2x + loge x x THINK WRITE a i f(x) = ecos 2x + loge x a i Differentiate ecos 2x by the chain 1 rule short cut and loge x by rule to f ′(x) = −2 sin 2x ecos 2x + --x obtain f ′(x). 1 ii 1 Express --- in index notation ii f ′(x) = −2 sin 2x ecos 2x + x−1 x so it can be differentiated. Differentiate f ′(x) to obtain f ′′(x). f ′′(x) = −4 cos 2x ecos 2x + 4 sin2 2x ecos 2x −x−2 2 1 Use the product rule to or = 4ecos 2x (sin2 2x − cos 2x) − ----2cos 2x x differentiate −2 sin 2x e . Continued over page

Chap 05 SM Page 178 Thursday, October 12, 2000 10:22 AM

178

Specialist Mathematics

THINK b i

ii

WRITE

1

Express

2

Express f(x) as a product.

3

Differentiate f(x) using the product rule to obtain f ′(x).

1

x in index notation.

b i

Differentiate f ′(x) using the product rule to obtain f ′′(x).

sin x f(x) = ----------x sin x f(x) = ---------1 --2 x = x

1 – --2

f ′(x) = − 1--2- x ii f ′′(x) = 3--4- x

Simplify by collecting like terms.

= 3--4- x

3

Simplify f ′′(x) by taking out

= 1--4- x

5 2

– ---

.

or

3 2

– ---

5 2

– ---

− 1--2- x

2

the factor 1--4- x

sin x

5 2

– ---

5 2

– ---

sin x − 1--2- x 3 2

– ---

1 2

– ---

sin x + x

3 2

– ---

cos x − x

sin x − x

cos x

1 – --2

3 2

– ---

cos x

sin x

cos x − x

1 2

– ---

sin x

(3 sin x − 4x cos x − 4x2 sin x)

3 sin x – 4x cos x – 4x 2 sin x ------------------------------------------------------------------5 --2 4x

WORKED Example 7 kx

-----dy d2 y If y = e 2 and --------2- – 3 ------ + 2 y = 0 , find the value of k. dx dx THINK

1

dy Find ------ from y = dx

kx ----e2 .

WRITE kx -----

y= e2

kx -----

dy ke 2 ------ = ---------2 dx 2

3

d2 y Find -------2- . dx dy d2 y Substitute ------ and -------2- into the equation dx dx 2 dy d y -------- − 3 ------ + 2y = 0. dx dx 2 kx -----

4

Take 1--4- e 2 out as a factor.

5

Factorise the quadratic function of k. kx 1 ------- e 2 4

(Note: cannot equal zero.) State the solutions. 6

kx -----

d2 y k 2e 2 -------2- = -----------4 dx kx ----2 k e2

kx -----

kx

----3ke 2 so ------------ – ------------- + 2e 2 = 0 4 2

kx 1 ------- e 2 (k2 4 kx 1 ------- e 2 4

− 6k + 8) = 0

(k − 4)(k − 2) = 0

Therefore k = 2 or k = 4.

Chap 05 SM Page 179 Thursday, October 12, 2000 10:22 AM

Chapter 5

Differential calculus

179

remember remember 2

d y 1. -------2- is the second derivative of y with respect to x. It is found by differentiating dx y twice. 2 d y 2. If y = f(x) then -------2- = f ′′(x). dx

5B WORKED

5

d2 y 1 Find -------2- if dx a y = 4x2 + 7x − 3 d y = x4 + 2x3 − 3x + 1 g y = 4x3 + 2 x – 1 2 j y = ------- + 3x –1 x

Math

b y = 5x − x + 3x 3

2

e y = x6 + 2x4 − 3x 9 --2

5 --3

h y = 2x – 6x + 5

c

cad

Example

Second derivatives

y = 6x 7 --2

3 --2

f

y= x +x

i

y = x2 + 5x−3

k y = 2x−2 + x−1

2 multiple choice If y = sin x , then: dy a ------ is equal to: dx x cos

A

x

1 D sin ---------2 x d2 y b -------2- is equal to: dx A x cos

x

1 D sin  – -------------  4x x WORKED

Example

6

cos x B --------------2 x 1 E cos ---------2 x

x sin x – cos x B ------------------------------------------x x – cos x – x sin x E -------------------------------------------------4x x

3 Find i f ′(x) and ii f ′′(x) if f(x) is equal to: a loge2x b x2 − loge x x d 2 sin --- + x 2 g tan 2x

4 --3

–cos x C -----------------x

x sin x – cos x C ------------------------------------------2x x

c

sin 2x + 4 cos x

e 2e5x − 3e x + 1

f

4e−3x + 3x3

h −3 tan (−4x) + 1

i

cos 2x --------------x

k e2 sin x

l

cos e5x

5 ---

j m

x2 -------------sin 2x sin 2x

5 ---

n ( cos 4x ) 2

o loge(cos x)

Second derivatives

Chap 05 SM Page 180 Thursday, October 12, 2000 10:22 AM

180

Specialist Mathematics

4 For each of the following functions f(x), match the graph which could represent f ′′(x). a f (x) = x3 + 2x b f (x) = 4x2 c f (x) = x4 − x2 d f (x) = sin x e f (x) = loge x f f (x) = e2x g f (x) = 4 x A B C y y y f"(x) 0

D

f"(x)

x

0

E

y

x f"(x)

F

y

x

0

y

f"(x)

0

G

y

f"(x)

x

x

0

H

f"(x)

f"(x)

I

y

x

0

y f"(x)

0

x

x

0

0

f"(x)

d2 y 5 If y = cos x, show that -------2- = – y . dx d2 y 6 If y = sin 3x, show that: -------2- + 9y = 0. dx dy d2 y 7 If y = x loge x, show that: ------ = ( x + y ) × -------2- . dx dx d2 y dy y 8 If y = xe x, show that: -------2- = ------ + -- . dx x dx d2 y WORKED 9 Find the value of k if y = e−kx and -------2- = 25y . Example dx 7 dy d2 y kx 10 If y = e and -------2- – 3 ------ – 4y = 0, find the value of k. dx dx 11 The position of a particle travelling in a straight line is given by the equation: x(t) = t3 − t2 − t + 7, where x has units in cm and t is in seconds. Find: dx a ------ , that is, the velocity at any time t dt d2 x b -------2- , that is, the acceleration at any time t dt dx c when and where the particle momentarily stops; that is, when ------ = 0 dt d the minimum velocity.

x

Chap 05 SM Page 181 Thursday, October 12, 2000 10:22 AM

Chapter 5

Differential calculus

181

Analysing the behaviour of functions using the second derivative The second derivative of a function can be used for testing the nature of stationary points, but is not a requirement of this course. It is provided as an alternative, or extension, to the first derivative test which is a requirement to Mathematical Methods Units 3 and 4.

First derivative function dy We have seen how the first derivative of a function, f ′(x) or ------ , can tell us where a dx function has a positive gradient, a negative gradient or a zero gradient (stationary point). For example, let us look at the functions f(x) = x2 and f(x) = x3.

Function f(x) = x2

Examine the graph at right. Since f (x) = x2, y f(x) = x2 f ′(x) = 2x x0 and f ′(x) = 0 at x = 0 f (x) is f(x) is 1. If f ′(x) = 0 decreasing increasing x At x = 0, f (x) is a stationary point. 0 x = 0 2. If f ′(x) < 0 f (x) is neither This occurs when 2x < 0. So if x < 0, then f (x) is increasing nor a decreasing function, one with a negative decreasing gradient for all x < 0. 3. If f ′(x) > 0 This occurs when 2x > 0. So if x > 0, then f (x) is an increasing function, one with a positive gradient for all x > 0. Consequently, at x = 0, a minimum stationary point occurs.

Function f(x) = x3

y

Examine the graph at right. Since f (x) = x , f ′(x) = 3x2 and f ′(x) = 0 at x = 0 1. If f ′(x) = 0 At x = 0, f (x) is a stationary point. 2. Here, f ′(x) > 0 for all values of x except zero. Consequently, at x = 0, a stationary point of inflection occurs. 3

Second derivative function

(

x0 f(x) is increasing x x=0 f(x) is neither increasing nor decreasing

)

d2 y Similarly the second derivative f ′′(x) or -------2- can tell us where the gradient function dx

(

)

dy f ′(x) or ------ is increasing or decreasing or neither (that is, changing from increasing dx to decreasing or vice-versa). Let us look at the situation when f ′′(x) is greater than, less than and equal to zero.

Chap 05 SM Page 182 Thursday, October 12, 2000 10:22 AM

Specialist Mathematics

Function f ′′(x) > 0

f(x) = x2

y Examine the graph at right. When f ′′(x) > 0 then the f ' (3) = 6 f ' (–3) = 6 gradient function f (x) is increasing. That is, as x increases then f ′(x) increases. f ' (2) = 4 f ' (–2) = –4 Again consider: f (x) = x2 f ' (1) = 2 f ' (–1) = –2 f ′(x) = 2x x 0 f ′′(x) = 2 –3 –2 –1 1 2 3 Gradient is always increasing from left to right

Function f ′′(x) < 0

Examine the graph at right. When f ′′(x) < 0, then the y f ' (0) = 0 gradient of f(x) is decreasing. That is, as x increases f ' (–1) = 2 f ' (1) = –2 then f ′(x) decreases. x 0 2 3 –3 –2 –1 1 For example, if f ' (2) = –4 f ' (–2) = 4 f(x) = −x2 f ′(x) = −2x (therefore there is a stationary f ' (3) = –6 point at x = 0) f ' (–3) = 6 f(x) = –x2 f ′′(x) = −2 Gradient is always decreasing that is, f ′′(x) < 0 for all values of x. from left to right This means that the gradient function f ′(x) is always decreasing.

Function f ′′(x) = 0 The gradient of f (x) is changing from increasing to f (x) = x3 y decreasing or vice-versa (providing the sign of f ′′(x) f ' (2) = 12 changes at the point where f ′′(x) = 0). Again consider: f(x) = x3 f ' (1) = 3 f ' (–1) = 3 f ′(x) = 3x2 (therefore there is a stationary point x at x = 0) –2 –1 0 1 2 f ′′(x) = 6x x>0 f ' (–2) = 12 f ' (x) is increasing so f ′′(x) = 0 at x = 0 and f ′′(x) < 0 if x < 0, x 0.

f ′(x) > 0 if x > 1.

3

Find when f ′(x) < 0.

f ′(x) < 0 if x < 1.

4

Sketch f ′(x).

y f'(x)

0

5

Find where f (x) is smooth and continuous and hence find the domain of f ′(x).

Domain is R.

1

x

264

Mathematical Methods Units 3 and 4

Gradient function of cubic functions The gradient function of a cubic is a quadratic function.

WORKED Example 3 Sketch the gradient function of the following function and state its domain.

y f (x)

THINK

1

WRITE

1

Find when f ′(x) = 0.

f ′(x) = 0 if x = −3 and x = 1.

2

Find when f ′(x) > 0.

f ′(x) > 0 if x < −3 and x > 1.

3

Find when f ′(x) < 0.

f ′(x) < 0 if −3 < x < 1.

4

Sketch the graph of the gradient function.

y

–3

5

Find the domain by determining where f (x) is smooth and continuous.

x

0

–3

0

f'(x)

1

x

Domain is R.

WORKED Example 4 For the function f(x) shown, state the domain of the gradient function f ′(x).

y

–1 0 2

x f(x)

THINK

WRITE

The function is smooth and continuous everywhere except at x = −1 (discontinuous) and x = 2 (not smooth).

Domain = R\{−1, 2}

Chapter 7 Differentiation

265

remember remember 1. A function is smooth if there are no sharp points on its graph. 2. A function is continuous if the graph can be drawn without lifting pen from paper. 3. The gradient of a function exists where the function is smooth and continuous. 4. The gradient function of a polynomial function is also a polynomial function but the degree is reduced by 1. For example, the gradient function of f (x) = ax2 + bx + c is of the form y = mx + c. 5. Wherever the gradient of a function is zero, its gradient function will have an x-intercept. 6. Wherever the gradient of a function is positive (sloping /), the gradient function graph is above the x-axis. 7. Wherever the gradient of a function is negative (sloping \), the gradient function graph is below the x-axis. 8. The gradient of a horizontal line is 0. 9. The gradient of a vertical line is undefined.

Example

1

1 For each straight line function shown below, sketch the graph of its gradient function. a

b

y

c

y

y HEET

f (x) 3

1

f(x) –2

x

0

–1

0

x

1 f (x)

d

y 0

e

f(x) 2

y f (x)

x 3

0 –5

HEET

SkillS

WORKED

x

0 –1

x

7.1

SkillS

7A

Graphing the gradient function from the graph of a function

7.2

266

Mathematical Methods Units 3 and 4

2 multiple choice a The gradient of the line in the graph at right is: A 1

B 2

f(x)

C 0 2

E −1

1 --2

D

y

x

0 1

b The graph of the gradient function in the graph at right is represented by which of the diagrams below? A

B

y

C

y

y

f'(x) f'(x)

2

D

0

x

0 1

E

y 1– 2

x

2

x

0

y x

0

f'(x)

f'(x)

2

1

f'(x) x

0

WORKED

Example

–2

3 Sketch the graph of the gradient function for each quadratic function shown below.

2

a

b

y

c

y

y

g(x) x

0

0 g(x)

x

0

–2

d

y

e

g(x)

y –2

0

3

x g(x)

g(x)

0 –1

x

x

267

Chapter 7 Differentiation

4 multiple choice a The gradient of the function shown in the graph at right is: A always increasing B always decreasing C decreasing then increasing D increasing then decreasing E constant

y

b The gradient function for the graph at right is shown by which of the graphs shown below?

0

A

B

y

y

x

0

f(x)

C

f'(x)

y

x

0

x

f'(x)

x

0

f'(x)

D

E

y

y

f'(x)

x

0

x

0 f'(x)

WORKED

Example

3

5 For each cubic function f (x) graphed below, sketch the gradient function. a

b

y

c

y

y

f (x)

–3

0

2

x

0

1

4

x

0

f(x)

5

x f (x)

d

e

y Gradient = 0

f(x)

0

x

f

y f(x)

y

Gradient = 0 0

x –3

0

x f (x)

g

f(x)

y Gradient = 0 2 0

x

268

Mathematical Methods Units 3 and 4

6 multiple choice a The figure at right has a positive gradient where:

y f(x)

A −1 < x < 2 B x < −1 only –1

C x > 2 only

0

x

2

D x < −1 and x > 2 E x>0 b The figure above has a negative gradient where: A x > −1 B x 2 E x 2 seconds the javelin is travelling downward. d

1 2 3

4 5

Find the time at which a height of 17 m occurs, by substituting h = 17 into h(t). Make RHS = 0. Solve for t. Note: the quadratic formula could also be used to solve for t. The first time it reaches 17 m is the smaller value of t. Evaluate h′(1).

d

20t − 5t2 + 2 = 17 −5t2 + 20t − 15 = 0 t 2 − 4t + 3 = 0 (t − 1)(t − 3) = 0 t = 1 or 3 It first reaches 17 m when t = 1 s. h′(1) = 20 − 5(1)2 = 15 m/s Rate of change of height is 15 m/s.

It is worth noting that there are two common ways of writing the derivative as a function. For example, the derivative of the function P(x) = x2 + 5x − 7 may be written as dP P′(x) = 2x + 5 or as ------- = 2x + 5 . dx

398

Mathematical Methods Units 1 and 2

WORKED Example 3 The shockwave from a nuclear blast spreads out at ground level in a circular manner. a Write down a relationship between the area of ground, A km2, over which the shockwave passes and its radius, r km. b Find the rate of change of A with respect to r. c Find the rate of change of A when the radius is 2 km. d What is the rate of change of A when the area covered is 314 km2?

r km Area A km2

THINK

WRITE

a State the formula for the area of a circle.

a A(r) = π r 2

b Differentiate A(r).

b A′(r) = 2π r

c Substitute r = 2 into A′(r). Note: The units for the rate of change of A (km2) with respect to r (km) are km2 per km or km2/km.

c A′(2) = 2(3.14)(2) = 12.56 Rate of change of A when the radius is 2 km is 12.56 km2/km.

d

1

Substitute A = 314 into the area function A(r) and solve for r.

2

Find the rate of change when r = 10.

d A(r) = π r 2 314 = 3.14r 2 314 r 2 = ---------3.14 = 100 r = 10 since r > 0 A′(10) = 2π (10) = 62.8 Rate of change of A when area is 314 km2 is 62.8 km2/km.

remember remember Average rates of change are calculated using the original function, while differentiation of this function is needed in order to calculate instantaneous rates of change at specific points.

Chapter 9 Applications of differentiation

9A Example

1

Rates of change

1 If f (x) = x2 + 5x + 15 find: a the average rate of change between x = 3 and x = 5 b a new function that describes the rate of change c the (instantaneous) rate of change when x = 5.

HEET

2 A balloon is inflated so that its volume, V cm3, at any time, t seconds later is:

3 multiple choice

Gradient between two points on a graph E

L Spread XCE

Gradient between two points on a graph

The average rate of change between x = 1 and x = 3 for the function y = x2 + 3x + 5 is: A 1 B 9 C 5 D 3 E 7 4 multiple choice

The instantaneous rate of change of the function f (x) = x3 − 3x2 + 4x, when x = −2 is: A 2 B −2 C 28 D 3 E 12 5 multiple choice dy If the rate of change of a function is described by ------ = 2x2 − 7x, then the function dx could be: C y = 2--3- x3 − 7--2- x2 + 5 B y = 2--3- x3 − 7x A y = 6x3 − 14x D y = x3 − 7--2- x2 + 2 WORKED

Example

2

E 2x2 − 7x + 5

6 In a baseball game the ball is hit so that its height above the ground, h metres, is h(t) = 1 + 18t − 3t 2 t seconds after being struck. a Find the rate of change, h′(t). b Calculate the rate of change of height after: i 2 seconds ii 3 seconds iii 4 seconds. c What happens when t = 3 seconds? d Find the rate of change of height when the ball first reaches a height of 16 metres.

sheet

a What is the volume of the balloon when: i t = 0? ii t = 10? b Hence, find the average rate of change between t = 0 and t = 10. c Find the rate of change of volume when i t=0 ii t = 5 iii t = 10.

Math

cad

V = − 8--5- t 3 + 24t 2, t ∈ [0, 10]

9.1

SkillS

WORKED

399

400

SkillS

HEET

9.2

Mathematical Methods Units 1 and 2

7 The position, x metres, of a lift (above ground level) at any time, t seconds, is given by: x(t) = −2t2 + 40t a Find the rate of change of displacement (velocity) at any time, t. b Find the rate of change when: i t=5 ii t = 9 iii t = 11. c What happened between t = 9 and t = 11? d When and where is the rate of change zero? 8 The number of seats, N, occupied in a soccer stadium t hours after the gates are opened is given by: N = 500t 2 + 3500t, t ∈ [0, 5] a Find N when: i t = 1 and ii t = 3. b What is the average rate of change between t = 1 and t = 3? c Find the instantaneous rate when: i t=0 ii t = 1 iii t = 3 iv t = 4. d Why is the rate increasing in the first 4 hours? 9 The weight, W kg, of a foal at any time, t weeks, after birth is given by: 3 2 -t W = 80 + 12t − ----where 0 < t < 20 10 a What is the weight of the foal at birth? b Find an expression for the rate of change of weight at any time, t. c Find the rate of change after: i 5 weeks ii 10 weeks iii 15 weeks. d Is the rate of change of the foal’s weight increasing or decreasing? e When does the foal weigh 200 kg?

Chapter 9 Applications of differentiation

401 3 ---

10 The weekly profit, P (hundreds of dollars), of a factory is given by P = 4.5n – n 2 , where n is the number of employees. dP a Find ------- . dn b Hence, find the rate of change of profit, in dollars per employee, if the number of employees is: i 4 ii 16 iii 25. c Find n when the rate of change is zero. 11 Gas is escaping from a cylinder so that its volume, V cm3, t seconds after the leak 1 2 -t . starts, is described by V = 2000 − 20t − -------100 a Find the rate of change after: i 10 seconds ii 50 seconds iii 100 seconds. b Is the rate of change ever positive? Why? WORKED

Example

3

12 Assume an oil spill from an oil tanker is circular and remains that way. a Write down a relationship between the area of the spill, A m2, and the radius, r metres. b Find the rate of change of A with respect to the radius, r. c Find the rate of change of A when the radius is: i 10 m ii 50 m iii 100 m. d Is the area increasing more rapidly as the radius increases? Why? 13 A spherical balloon is being inflated. a Express the volume of the balloon, V m3, as a function of the radius, r metres. b Find the rate of change of V with respect to r. c Find the rate of change when the radius is: i 0.1 m ii 0.2 m iii 0.3 m.

r

402

Mathematical Methods Units 1 and 2

14 A rectangular fish tank has a square base with its height being equal to half its base length. a Express the length and width of the base in terms of its height, h. b Hence, express the volume, V m3, in terms of the height, h, only. c Find the rate of change of V when: i h=1m ii h = 2 m iii h = 3 m. 15 For the triangular package shown find: a x in terms of h b the volume, V, as a function of h only c the rate of change of V when i h = 0.5 m ii h = 1 m.

6

30° x

h

16 A new estate is to be established on the side of a hill. y

80

200

x

Regulations will not allow houses to be built on slopes where the gradient is greater than 0.45. If the equation of the cross-section of the hill is y = −0.000 02x3 + 0.006x2 find: dy a the gradient of the slope -----dx b the gradient of the slope when x equals: i 160 ii 100 iii 40 iv 20 c the values of x where the gradient is 0.45 d the range of heights for which houses cannot be built on the hill. 17 A bushfire burns out A hectares of land, t hours after it started according to the rule A = 90t 2 − 3t 3 a At what rate, in hectares per hour, is the fire spreading at any time, t? b What is the rate when t equals: i 0 ii 4 iii 8 iv 10 v 12 vi 16 vii 20? c Briefly explain how the rate of burning changes during the first 20 hours. d Why isn’t there a negative rate of change in the first 20 hours? e What happens after 20 hours? f After how long is the rate of change equal to 756 hectares per hour?

Chapter 9 Applications of differentiation

403

Sketching graphs containing stationary points The derivative of a function gives its gradient function — that is, it gives the gradient of a tangent to the curve for any specified value of the independent variable. When the derivative equals zero the tangent is horizontal. The point, or points, on the curve where this occurs are called stationary points. In other words, a function f (x) has stationary points when f ′(x) = 0. Stationary points can take the form of: 1. A local minimum turning point 2. A local maximum turning point, or 3. A point of inflection.

Local minimum turning point

y

Just to the left of a, the gradient is negative; that is, if x < a, but close to a, then f ′(x) < 0. At the point where x = a the gradient is zero; that is, at f'(x) < 0 x = a, f ′(x) = 0. Just to the right of a the gradient is positive; that is, if x > a, but close to a, f ′(x) > 0. 0 In other words, for a stationary point at x = a, if the gradient changes from negative to positive as we move from left to right in the vicinity of a, it is a local minimum.

Local maximum turning point

f'(x) > 0 f'(a) = 0 x

a

y

If x < a, but close to a, then f ′(x) > 0. f'(a) = 0 At x = a, f ′(x) = 0. If x > a, but close to a, f ′(x) < 0. f'(x) < 0 In other words, for a stationary point at x = a. If the f'(x) > 0 gradient changes from positive to negative as we move x 0 from left to right in the vicinity of a, it is a local maximum. a The term local maximum or local minimum implies that the function has a maximum or minimum in the vicinity of x = a. This is important because some functions can have more than one stationary point.

Point of inflection y

y f'(x) > 0 f'(x) < 0 f'(a) = 0

f'(a) = 0

f'(x) > 0 0

a

x

f'(x) < 0 0

a

x

If x < a, but close to a, then f ′(x) > 0. If x < a, but close to a, f ′(x) < 0. At x = a, f ′(x) = 0. At x = a, f ′(x) = 0. If x > a, but close to a, f ′(x) > 0. If x > a, but close to a, f ′(x) < 0. In other words, for a stationary point at x = a, if the gradient remains positive or negative in the vicinity of a it is a point of inflection.

404

Mathematical Methods Units 1 and 2

WORKED Example 4

If f (x) = x 3 − 6x 2 − 15x, find: a the value(s) of x where the gradient is zero

b the stationary point(s).

THINK

WRITE

a

Write the function. Differentiate f (x) to find the gradient function f ′(x). Solve f ′(x) = 0 to find the x-values of each stationary point.

a f(x) = x3 − 6x2 − 15x f ′(x) = 3x2 − 12x − 15

1

Substitute each value of x into f (x) to find the corresponding y-values.

b

2

Write the coordinates of each stationary point.

1 2 3

b

For stationary points: f ′(x) = 0 3x2 − 12x − 15 = 0 3(x2 − 4x − 5) = 0 x2 − 4x − 5 = 0 (x − 5)(x + 1) = 0 x = 5 or x = −1 f (5) = (5)3 − 6(5)2 − 15(5) = −100 f (−1) = (–1)3 − 6(−1)2 − 15(−1) =8 Stationary points occur at (5, −100) and (−1, 8).

WORKED Example 5

If y = 1--3- x3 + 2x2 + 3x − 2 find all stationary points and determine their nature. THINK 1 2 3

4

WRITE

Write the rule.

y = 1--3- x3 + 2x2 + 3x − 2

dy Differentiate y to find the gradient function ------ . dx dy Solve ------ = 0 to find the x-values of dx each stationary point. Substitute each value of x into

dy ------ = x2 + 4x + 3 dx dy For stationary points: ------ = 0 dx x2 + 4x + 3 = 0 (x + 3)(x + 1) = 0 x = −3 or x = −1 When x = −3,

y = 1--3- x3 + 2x2 + 3x − 2 to find the

y = 1--3- (−3)3 + 2(−3)2 + 3(−3) – 2

corresponding y-values.

y = −2 One stationary point is (−3, −2). When x = −1, y = 1--3- (−1)3 + 2(−1)2 + 3(−1) − 2 y = −3 1--3The other one is (−1, −3 1--3- ).

Chapter 9 Applications of differentiation

THINK 5

Determine the nature of the stationary dy point at x = −3 by evaluating ------ to the left dx and right, say at x = −4 and at x = −2.

6

Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = −3, the stationary point (−3, −2) is a local maximum.

WRITE When x = −4 dy ------ = (−4)2 + 4(−4) + 3 dx = 3. When x = −2 dy ------ = (−2)2 + 4(−2) + 3 dx = −1. Zero gradient

Determine the nature of the stationary dy point at x = −1 by evaluating ------ to the left dx and right, say at x = −2 and at x = 0.

8

Since the gradient changes from negative to positive as we move from left to right in the vicinity of x = −1, the stationary point (−1, −3 1--3- ) is a local minimum.

Negative gradient

Positive gradient

–4 7

405

–3

x

–2

(−3, −2) is a local maximum. When x = −2 dy ------ = (−2)2 + 4(−2) + 3 dx = −1. When x = 0 dy ------ = (0)2 + 4(0) + 3 dx = 3. Positive gradient

Negative gradient Zero gradient

–2

–1

x

0

(−1, −3 1--3- ) is a local minimum.

WORKED Example 6

Sketch the graph of the function f (x) = 5 + 4x − x2 labelling all intercepts and stationary points. THINK 1 2

Write the function. Find the y-intercept by letting x = 0.

WRITE f(x) = 5 + 4x − x2 y-intercept: if x = 0, f (0) = 5 + 4(0) − (0)2 =5 so y-intercept is (0, 5). Continued over page

406

Mathematical Methods Units 1 and 2

THINK 3

4 5

6

7

8

WRITE

Find the x-intercepts by letting f (x) = 0.

Differentiate f (x) to find the gradient function f ′(x). Solve f ′(x) = 0 to find the x-value/s of each stationary point.

Substitute this value of x into f (x) = 5 + 4x − x2 to find the corresponding y-value. Determine the nature of the stationary point at x = 2 by evaluating f ′(x) to the left and right, say at x = 1 and x = 3. Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 2, the stationary point (2, 9) is a local maximum.

x-intercepts: if f(x) = 0, 5 + 4x − x2 = 0 x2 − 4x − 5 = 0 (x + 1)(x − 5) = 0 x = −1 or x = 5 so x-intercepts are (−1, 0) and (5, 0). f ′(x) = 4 − 2x For stationary points: f ′(x) = 0 4 − 2x = 0 −2x = −4 x=2 f (2) = 5 + 4(2) − (2)2 =9 so there is a stationary point at (2, 9). f ′(1) = 4 − 2(1) =2 f ′(3) = 4 − 2(3) = −2 Zero gradient Positive gradient

Negative gradient

1

2

3

x

(2, 9) is a local maximum. 9

Sketch the graph.

y (2, 9) 9 f (x)

5

–1 0

2

5 x

Graphics Calculator tip! Finding stationary (turning) points To find a local maximum or minimum point on a function, enter it in the Y= menu and follow these steps. 1. Press 2nd [CALC] and select 4:maximum (or 3:minimum). 2. Use the arrow key to scroll to the left of a turning point and press ENTER . 3. Use the arrow key to scroll to the right of the same turning point and press ENTER twice.

Chapter 9 Applications of differentiation

407

remember remember When sketching graphs of polynomial functions, four main features should be indicated: 1. the y-intercept (found by calculating y when x = 0) 2. the x-intercept(s) (found by solving the equation for x when y = 0) dy 3. the stationary point(s) (found by solving the equation ------ = 0) dx 4. the type of stationary point(s) (found by checking the sign of the gradient to the left and right of the stationary point).

9B WORKED

Example

2 For each function in question 1 determine all of the stationary points.

4b

4 For the function f (x) = 5 − x2: a find x when f ′(x) = 0

L Spread XCE

sheet

3 If f (x) = x2 − 8x + 1: a show that it has a stationary point when x = 4 b evaluate f ′(3) and f ′(5) c state which type of stationary point it is.

Quadratic graphs

E

WORKED

sheet

E

4a

Math 1 For each of the following functions determine the value(s) of x where the gradient is zero. Quadratic a f (x) = x2 + 2x b f (x) = x2 − 8x + 5 graphs 3 2 3 2 c f (x) = x − 3x d f (x) = 2x + 6x − 18x + 1 e y = (x + 6)(x − 2) f y = x2(x − 1) 2 L Spread g y = 10 + 4x − x h y = 1--3- x3 − 3x2 + 5x − 2 XCE

cad

Example

Sketching graphs containing stationary points

Cubic graphs

b state what type of stationary point it has

6 If f (x) = 1--3- x3 − x2 − 3x + 5: a show there are stationary points when x = −1 and x = 3 c state what type of stationary points they are.

b evaluate f ′(−2), f ′(0) and f ′(4)

7 multiple choice When x = 1, the curve y = 2x2 − 3x + 1: A is decreasing B has a local maximum D is increasing E does not exist

C has a local minimum

8 multiple choice When x = −1 the function y = x3 − 2x2 − 7x: A is decreasing B has a local maximum D is increasing E does not exist

C has a local minimum

Math

cad

5 If f (x) = x3 − 4 then: a show that it has a stationary point when x = 0 only b find f ′(−1) and f ′(1) c state which type of stationary point it is

Cubic graphs

408

Mathematical Methods Units 1 and 2

9 multiple choice The graph below best representing a function with f ′(−2) = 0, f ′(x) < 0 if x < −2, and f ′(x) > 0 if x > −2 is: A B C y y y

f (x)

–2

x

0

f (x)

f (x)

D

–2

E

y

y

f (x)

x

0

x

0

f (x)

x

0 –2

–2

x

0

–2

10 multiple choice f ′(1) = f ′(4) = 0 and f ′(x) < 0 if 1 < x < 4 and f ′(x) > 0 if x < 1 and x > 4. The graph below which satisfies these conditions is: A B C y y y f (x) f (x) f (x) 0 0

D

1

1

4

x

x

4

E

y

0

1

4

x

y f (x)

0

1

4

x

0

1

4

x

f (x) WORKED

Example

5

9.3

SkillS

HEET

WORKED

Example

SkillS

HEET

9.4

6

11 For each of the following find the stationary points and determine their nature. b y = 8x − 2x2 c y = x3 − x2 a y = x2 + 6x + 2 1 2 1 3 1 2 3 d y = x + --2- x − 3 e y = --3- x − --2- x − 2x f y = (x − 1)3 3 3 g y=x +3 h y = x − 27x + 5 12 Sketch graphs of the following functions, labelling all intercepts and stationary points. b f (x) = x3 − 3x − 2 a f (x) = x2 − 2x − 3 3 2 c f (x) = x − 2x + x d f (x) = x2(3 − x) 3 2 e f (x) = x + 4x + 4x f f (x) = x3 − 4x2 − 11x + 30 3 g f (x) = (x + 2) h f (x) = 24 + 10x − 3x2 − x3 3 2 i f (x) = x − 2x − x + 2 j f (x) = 8 − x3

Chapter 9 Applications of differentiation

409

Solving maximum and minimum problems There are many practical situations where it is necessary to determine the maximum or minimum value of a function. For quadratic functions, differentiation makes this a relatively simple task because as we saw in the previous section, setting the derivative equal to zero allows us to solve an equation to obtain the value(s) of x for which the local maximum or minimum values (turning points) occur. When solving maximum or minimum problems it should be verified that it is in fact a maximum or minimum by checking the sign of the derivative to left and right of the turning point. y Local f (x) In the case of cubic and higher order polynomials, maximum Absolute the local maximum (or minimum) may or may not be maximum in the interval the highest (or lowest) value of the function in a given [a, b] domain. An example where the local maximum, found by 0 a b x solving f ′(x) = 0, is not the largest value of f(x) in the domain [a, b] is shown. Here, B is the point where f(x) is greatest in this domain, and is called the absolute maximum for the interval.

Case 1. The function is known

WORKED Example 7 A baseball fielder throws the ball so that the equation of its path is: y = 1.5 + x − 0.02x2 where x (metres) is the horizontal distance travelled by the ball and y (metres) is the vertical height reached. a Find the value of x for which the maximum height is reached (verify that it is a maximum). b Find the maximum height reached. THINK

WRITE

a

a y = 1.5 + x − 0.02x2 dy ------ = 1 − 0.04x dx

1 2 3

4

Write the equation of the path. dy Find the derivative ------ . dx dy Solve the equation ------ = 0 to find the dx value of x for which height is a maximum. Determine the nature of this stationary dy point at x = 25 by evaluating ------ to the dx left and right, say, at x = 24 and at x = 26.

dy For stationary points: ------ = 0 dx 1 − 0.04x = 0 −0.04x = −1 x = 25 When x = 24 dy ------ = 1 − 0.04(24) Zero gradient dx Positive = 0.04. gradient When x = 26 dy ------ = 1 − 0.04(26) dx = −0.04.

Negative gradient

Continued over page

410

Mathematical Methods Units 1 and 2

THINK 5

WRITE

Since the gradient changes from positive to negative as we move from left to right in the vicinity of x = 25, the stationary point is a local maximum.

b Substitute x = 25 into y = 1.5 + x − 0.02x2 to find the corresponding y-value (maximum height).

The stationary point is a local maximum.

b When x = 25, y = 1.5 + 25 − 0.02(25)2 = 14 So the maximum height reached is 14 m.

Case 2. The rule for the function is not given If the rule is not given directly then the following steps should be followed: 1. Draw a diagram if necessary and write an equation linking the given information. 2. Identify the quantity to be maximised or minimised. 3. Express this quantity as a function of one variable only (often this will be x). 4. Differentiate, set the derivative equal to zero, and solve. 5. Determine, in the case of more than one value, which one represents the maximum or minimum value. 6. For some functions, a maximum or minimum may occur at the extreme points of the domain so check these also. 7. Answer the question that is being asked. 8. Sketch a graph of the function if it helps to answer the question, noting any restrictions on the domain.

WORKED Example 8 A farmer wishes to fence off a rectangular paddock on a straight stretch of river so that only 3 sides of fencing are required. Find the largest possible area of the paddock if 240 metres of fencing is available. THINK 1

2

Draw a diagram to represent the situation, using labels to represent the variables for length and width and write an equation involving the given information.

WRITE River w Fence

Fence w

Let w = width l = length P = perimeter

Fence l

P = l + 2w = 240 Write a rule for the area, A, of the paddock A = l × w in terms of length, l, and width, w.

[1] [2]

Chapter 9 Applications of differentiation

THINK

WRITE

Express the length, l, of the rectangle in terms of the width, w, using equation [1].

l + 2w = 240 l = 240 − 2w

Express the quantity, A, as a function of one variable, w, by substituting [3] into [2].

Substituting [3] into [2]: A(w) = (240 − 2w)w = 240w − 2w2

Solve A ′(w) = 0.

A′(w) = 240 − 4w For stationary points: A′(w) = 0 240 − 4w = 0 240 = 4w w = 60

6

Test to see if the stationary point at w = 60 will produce a maximum or minimum value for the area by evaluating A ′(w) to the left and right, say, at w = 59 and at w = 61.

When w = 59 A′(59) = 240 − 4(59) = 4. When w = 61 A′(61) = 240 − 4(61) = −4.

7

Since the gradient changes from positive to negative as we move from left to right in the vicinity of w = 60, the stationary point is a local maximum.

3

4

5

8

411

Find the maximum area of the paddock by substituting w = 60 into the function for area.

[3]

Zero gradient Positive gradient

Negative gradient

The stationary point is a local maximum. The area of the paddock is a maximum when w = 60. A(60) = (240 − 2 × 60) × 60 = 7200 m2

remember remember Defining a function and setting its derivative equal to zero to form an equation helps to tell us when a local maximum or minimum occurs. The solution(s) must then be substituted into the original function to find the actual maximum or minimum value(s).

412

Mathematical Methods Units 1 and 2

9C WORKED

d hca

Mat

Example

7

Spreadshe

et

EXCEL

Quadratic graphs

Quadratic graphs

Mat

d hca

1 A golfer hits the ball so that the equation of its path is: y = 1.2 + x − 0.025x2 where x (metres) is the horizontal distance travelled by the ball and y (metres) is the vertical height reached. a Find the value of x for which the maximum height is reached (and verify that it is a maximum). b Find the maximum height reached. 2 If the volume of water, V litres, in a family’s hot water tank t minutes after the shower is turned on is given by the rule V = 200 − 1.2t 2 + 0.08t 3, where 0 ≤ t ≤ 15: a find the time when the volume is minimum (that is, the length of time the shower is on) b verify that it is a minimum by checking the sign of the derivative c find the minimum volume d find the value of t when the tank is full again.

Spreadshe

et

EXCEL

Cubic graphs

Solving maximum and minimum problems

Cubic graphs

GC p

am rogr

3 A ball is thrown into the air so that its height, h metres, above the ground at time t seconds after being thrown is given by the function: h(t) = 1 + 15t − 5t 2 a Find the greatest height reached by the ball and the value of t for which it occurs. b Verify that it is a maximum.

Maximum

GC p

am rogr

Minimum WORKED Example 8

4 A gardener wishes to fence off a rectangular vegetable patch against her back fence so that only 3 sides of new fencing are required. Find the largest possible area of the vegetable patch if she has 16 m of fencing material available. 5 The sum of two numbers is 16. a By letting one number be x, find an expression for the other number. b Find an expression for the product of the two numbers, P. c Hence, find the numbers if P is a maximum. d Verify that it is a maximum. 6 The rectangle at right has a perimeter of 20 cm. a If the width is x cm, find an expression for the length. b Write an expression for the area, A, in terms of x Width = x only. c Find the value of x required for maximum area. d Find the dimensions of the rectangle for maximum area. e Hence, find the maximum area.

Length

Chapter 9 Applications of differentiation

413

7 A farmer wishes to create a rectangular pen to contain as much area as possible using 60 metres of fencing. a Write expressions for the dimensions (length and width) of the pen. b Hence, find the maximum area. 8 The cost of producing a particular toaster is $(250 + 1.2n2) where n is the number produced each day. If the toasters are sold for $60 each: a write an expression for the profit, P, dollars b find how many toasters should be produced each day for maximum profit c hence, find the maximum daily profit possible. 9 A company’s income each week is $(800 + 1000n − 20n2) where n is the number of employees. The company spends $760 per employee for wages and materials. a Write an expression for the company weekly profit, P dollars. b Determine the number of employees required for maximum profit and hence calculate the maximum weekly profit. 10 The sum of two numbers is 10. Find the numbers if the sum of their squares is to be a minimum. 11 A square has four equal squares cut out of the corners as shown at right. It is then folded to form an open rectangular box. a What is the range of possible values for x? b In terms of x find expressions for the i height, ii length, and iii width of the box. c Write an expression for the volume, V (in terms of x only). d Find the maximum possible volume of the box.

12 cm x 12 cm

12 The base and sides of a shirt box are to be made from a rectangular sheet of cardboard measuring 50 cm × 40 cm. Find: a the dimensions of the box required for maximum volume b the maximum volume. (Give answers correct to 2 decimal places.) 13 The volume of the square-based box shown at right is 256 cm3. a Find h in terms of l. If the box has an open top find: h b the surface area, A, in terms of l only l c the dimensions of the box if the surface area is to be a minimum l d the minimum area. (Hint: 1--- = l –1.) l 14 A closed, square-based box of volume 1000 cm3 is to be constructed using the minimum amount of metal sheet possible. Find its dimensions.

ET SHE

Work

1 2 - v dollars per hour, 15 The cost of flying an aircraft on a 900 km journey is 1600 + -------100 where v is the speed of the aircraft in km/h. Find: a the cost, C dollars, of the journey if v = 300 km/h b the cost, C dollars, of the journey in terms of v. (Hint: time = distance ÷ speed.) c the most economical speed and minimum cost.

9.1

414

Mathematical Methods Units 1 and 2

When is a maximum not a maximum? Often finding a maximum (or minimum) of a function is achieved by setting the derivative equal to zero and solving for the independent variable (such as x). This value of x is then substituted into the original function to find the maximum (or minimum) value of the function. One danger of this method is that it assumes the local maximum or minimum is the absolute maximum or minimum (see page 409). x3 Consider finding the maximum value of the function f(x) = ----- − 2x2 + 3x + 2 3 between x = 0 and x = 6: f ′(x) = x2 – 4x + 3 f ′(x) = (x − 1)(x + 3) So x = 1 or x = 3 for a local maximum or minimum. A computer generated graph of f(x) reveals that the absolute maximum of f(x) in the interval [0, 6] is not at x = 1 or x = 3! 3

y 10

f (x) = x–3 – 2x2 + 3x + 2

5 0 –2 –5

2 4 6

x

–10

The absolute maximum of f(x) obviously occurs at x = 6, and equals 63 f(6) = ----- − 2(32) + 3(3) + 2 3 = 20 1 Find the absolute maximum of f(x) = x3 + 5x2 − 8x − 12 between x = 0 and x = 3. 2 Find the minimum of f(x) = −x3 + 4x2 + 11x − 30 in the interval [1, 5]. 3 The temperature T°C of a pottery classroom x minutes after the class has started is described by the function T(x) = 0.000 08x(x + 2)2 + 21. Sketch a graph of the temperature during a 50-minute class and determine when the classroom is hottest, and what the temperature is then.

Chapter 9 Applications of differentiation

415

Applications of antidifferentiation We will now consider the situation where the derivative of a function (the gradient function) is known but the original function is unknown. Finding the original function requires a process called antidifferentiation. If f ′(x) = x n, n ∈ N, then 1 f (x) = ------------ x n + 1 + c n+1 where c represents a constant. 1 This can be verified by differentiating ------------ x n + 1 + c . n+1 The result is x n. Similarly, if f ′(x) = ax n, a ∈ R, n ∈ N, then a f (x) = ------------ x n + 1 + c n+1 dy We saw previously that an alternative expression for the derivative was ------ . dx dy Likewise, if ------ = x n , then dx 1 y = ------------ x n + 1 + c n+1 The derivative of a function is also known as the gradient function and describes the rate of change of the function.

WORKED Example 9

Find the rule for the function f(x) if f ′(x) = 3 + 4x − x2 and f(0) = 7. THINK 1 2 3

4

WRITE

Write the given expression. Antidifferentiate f ′(x) to obtain the general rule for f (x). Substitute x = 0 and f (x) = 7 into f (x) and solve to find the value of the constant, c. Write the rule for f (x).

f ′(x) = 3 + 4x − x2 x3 f (x) = 3x + 2x2 − ----- + c 3 ( 0 )3 7 = 3(0) + 2(0)2 − ---------- + c 3 c=7 x3 f (x) = 3x + 2x2 − ----- + 7 3

The pieces of information used to find the value of the constant that is generated following antidifferentiation are called boundary conditions.

416

Mathematical Methods Units 1 and 2

WORKED Example 10

The rate of change of the volume, V cm3, of a balloon at any time, t seconds, after it is inflated beyond 6 litres is given by: dV ------- = 3t 2 – 8t + 1 t ∈ [0, 3] dt a Express V as a function of t. b What is the volume of the balloon when t = 1? THINK

WRITE

a

dV a ------- = 3t 2 − 8t + 1 dt V(t) = t 3 − 4t 2 + t + c

1

Write the given expression.

2

Find the general rule for volume by antidifferentiation. Find the value of the constant, c, by substituting t = 0 and V = 6 into V(t). Write the rule for V(t).

3 4

b Substitute t = 1 into the volume function V(t).

6 = (0)3 − 4(0)2 + (0) + c c= 6 V(t) = t 3 − 4t 2 + t + 6 b V(1) = (1)3 − 4(1)2 + (1) + 6 =4 So the volume of the balloon at t = 1 is 4 litres.

remember remember When finding the antiderivative of a function: For each term in the function, increase the power of the variable by one and divide by the resulting power. Add a constant.

The Mathcad file ‘Antidifferentiation’ may be used to check your antiderivatives, and takes into account boundary conditions if there are any.

Mat

d hca

Antidifferentiation

Ch 09 MM 1&2 YR 11 Page 417 Friday, June 29, 2001 11:28 AM

Chapter 9 Applications of differentiation

9D WORKED

Example

9

417

Applications of antidifferentiation

1 Find the rule for the function f ( x) if f ′( x) = 3x2 − 2x and f (2) = 0.

Antidifferentiation

4 multiple choice If the gradient function of a curve that passes through the point (2, 2) is f ′( x) = 2x − 5, then the function f ( x) is: A x2 − 5x + 8 B x2 − 5x − 1 C x2 − 5 2 2 D x − 5x E x −2 5 multiple choice If f ′( x) = 4x + 1 and the y-intercept is −3 then f ( x) equals: A x2 + 2x − 3 B 2x2 + x − 1 C 2x2 + x − 3 2 2 D 2x + 2 x − 1 E x +x 6 multiple choice A curve passes through the point (2, 1) and has a gradient function f ′( x) = x(3x − 5). The function must be: A f (x) = x3 − 3x2 + 5 B f (x) = x3 − 5--2- x2 + 2 C f (x) = 3x2 − 5x − 1 D f (x) = x3 − 5--2- x2 + 3 WORKED

Example

10

E f (x) = 3--4- x4 − 5--2- x3 + 9

7 The velocity (v) of an aircraft is changing as it accelerates. Its acceleration (rate of change of velocity) at any time, t, after it begins accelerating from rest along a runway dv ------ = 6t 2 – 4t + 5 is given by dt where v is in km/h and t is in seconds. a Express v as a function of t. b Find the velocity after 5 seconds.

Math

cad

2 If f ′( x) = 3 + 5x − 2x2 and the y-intercept is 7, find f ( x). dy 3 The y-intercept of a curve is 10 and ------ = ( x + 1 ) ( x – 3 ) . Find the value of y when dx x = 3.

418

Mathematical Methods Units 1 and 2

8 The rate of change of position (velocity) of a particle travelling in a straight line is dx given by ------ = t 2 – 6t + 2 , where x is in metres and t is in seconds. dt If the particle starts at x = 1, find its position when t = 3. 9 The rate of increase of volume per unit increase in depth for a particular container is given by: dV ------- = 2 ( h + 5 ) 2 dh where V cm3 is the volume and the depth is h cm. a If V = 0 when h = 0, express V as a function of h. b Find the volume at a height of 7 cm. 10 The weekly rate of change of profit with respect to the number of employees, n, in a factory is: dP 3 ------- = 3.182 – --- n dn 4 where P is in thousands of dollars. a Find the number of employees for maximum profit (assume P = 0 when n = 0). b Hence find the maximum profit. 11 The rate of deflection from the horizontal of a 2 m long diving board when a 70 kg person is x metres from its fixed end is: dy ------ = –0.06 ( x + 1 ) 2 + 0.06 dx y

0

x Deflection

a What is the deflection, y, when x = 0. b Find the equation that measures the deflection at any point on the board. c Find the maximum deflection. (Be careful.)

Work

ET SHE

9.2

12 The rate of change of height of a hot dh air balloon is given by ------ = 4t − 1 dt where h is the height above the ground in metres after t seconds. a Write h as a function of t. b Find the height after 4 seconds. c How long does it take the balloon to reach a height of 60 metres?

h

Chapter 9 Applications of differentiation

419

summary Rates of change change in y • Average rate of change = --------------------------- . change in x dy • The derivative of a function, f ′(x) (or ------ ), is needed in order to calculate the dx (instantaneous) rate of change at a particular point. The rate of change of a function, f (x), at x = a is given by f ′(a).

Sketching graphs containing stationary points • Stationary points occur when f ′(x) = 0. • Three types of stationary point exist and, by testing the sign of y the gradient to the left and right of a stationary point, the nature (type) can be determined. 1. Local maximum turning points. ( f ′(x) changes from + to − moving left to right.)

0

Local maximum

x

y

2. Local minimum turning points. ( f ′(x) changes from − to + moving left to right.)

Local minimum

x

0 y

3. Points of inflection. ( f ′(x) remains the same sign on both sides moving left to right.)

Solving maximum and minimum problems

Point of inflection

0

x

• By solving the equation f ′(x) = 0, and substituting the solutions into the original function, the maximum or minimum value of a quantity may be found. When the function is not provided it is necessary to formulate a rule in terms of one variable using the information given. Drawing a diagram to represent the situation is often useful. • Always test to determine if a stationary point is a y Local f (x) maximum or a minimum by checking the sign of the maximum Absolute gradient to the left and right of the point. maximum in the interval • Check whether or not the local maximum or minimum [a, b] is the absolute maximum or minimum. The absolute 0 a b x maximum or minimum may be the value of the function at one end of a specified interval.

Applications of antidifferentiation • When the derivative of a function is known, antidifferentiation can provide the original function. Since the original function may have contained a ‘constant’, this must be allowed for, and can be found using the boundary conditions provided in the question.

420

Mathematical Methods Units 1 and 2

CHAPTER review Multiple choice

9A

1 The rate of change of f (x) = 2x3 − 5x2 + 7 when x = 2 is: A −4 B 7 C −36 D 0

9A

2 If V = −3t 2 + 7t + 50 then the average rate of change between t = 1 and t = 4 is: A −10 2--3B −10 C −6 D −8 E 0

9B

3 If f (x) = 5 + 15x + 6x2 − x3 then the gradient is zero when x equals: A 1 or −5 B 1 or 5 C −1 or 5 D −1 or −5

9B

4 The curve y = x2 − 10x + 21 has: A a local maximum at (5, 0) C a point of inflection at (5, 0) E a point of inflection at (5, −4)

9B

5 When x = −2, the graph of y = 2x2 + 3x − 5: A is increasing B has a local maximum C has a point of inflection D has a local minimum E is decreasing

9B

6 For a particular function g(x), g(1) = 0 and g′(x) < 0 if x ≥ 1. The graph which could represent g(x) is: A B C y y y g(x) g(x)

0

1

D

x

1

x

1

x

0

E

y

E 0 and −1

B a local minimum at (5, −4) D a local maximum at (5, −4)

g(x)

0

E 4

y

1 0

1

x

x

0

g(x)

g(x)

9B

7 The maximum value of f (x) = −2x2 + 8x is: A 40 B 0 C 4

9B

8 The local minimum value of h(x) = 1--3- x3 + 6x2 − 28x − 3 occurs when x equals: A 2 B –4 C 0 D −3 E 1

D −24

E 8

Chapter 9 Applications of differentiation

9 The function g(x) = (x + 3)3 has: A a local maximum when x = −3 C a local minimum where x = −3 E a point of inflection where x = 3

421

B a point of inflection when x = −3 D a local minimum where x = 3

10 A curve with a local maximum and a local minimum is: A y = x3 + 2x2 − 7x + 1 D y = (x − 2)3

B y = x2 − 3x + 1 E y = x2 + 6x

C y = x3 + 7

11 The antiderivative of 12x + 3 is: A 6x2 + 3x D 6x2 + 3x + c

B 24x2 + 3x + c E 6x2

C 24x2 + 3x

dy 12 If the gradient of a curve is ------ = ( x – 2 ) ( x + 5 ) and its y-intercept is −3, then its rule is: dx A y = 1--3- x3 + 3--2- x2 − 10x − 3

B y = 1--3- x3 + 3--2- x2 − 5x − 3

9B 9C 9D 9D

C y = x3 + 3x2 − 10x − 3

D y = 1--3- x3 + 3--2- x2 − 10x − 10 E y = 1--4- x4 − 10x2

Short answer 1 If the position of a particle moving in a straight line is given by the rule x(t) = −2t 2 + 8t + 3, where x is in centimetres and t is in seconds, find: a the initial position of the particle b the rate of change of displacement (that is, the velocity) at any time, t c the rate of change when t = 4 d when and where the velocity is zero e whether the particle is moving to the left or to the right when t = 3 f the distance travelled in the first 3 seconds.

9A

2 For the function f (x) = x3 − 3x + 2: a find the y-intercept b find the x-intercepts c find the stationary points and state their type d sketch the graph of f (x).

9B

3 If the volume of liquid in a vat, V litres, during a manufacturing process is given by V = 6t − t 2, where t ∈ [0, 6], find: a the rate of change 2 hours after the vat starts to fill up b when the vat has a maximum volume.

9C

4 If a piece of wire is 80 cm long: a find the area of the largest rectangle that can be formed by the wire b determine whether a circle would give a larger area.

9C

5 Find the maximum possible volume of a fully enclosed cylindrical water tank given that the total internal surface area of the tank is 600 π square units.

9C

422

Mathematical Methods Units 1 and 2

h

9D

6 The rate of increase of height, h metres, of an ascending helicopter at any time, t minutes after dh it takes off is ------ = t 2 – 14t + 45. dt a Find an expression for the height at any time. b Find the height 6 minutes after takeoff. c Find the maximum height reached in the first 9 minutes.

9D

7 A particle travels such that its velocity at any time, t, is given by v = 2t + 1. a Given that velocity represents the rate of change of position, x, write down the relationship between v and x. b If x = 3 when t = 2 write an expression for x in terms of t. c Find the position of the particle when t = 10.

Analysis

9B

1 A ball is thrown vertically up so that its height above the ground, h metres, at any time, t seconds, after leaving the thrower’s hand is given by the function h(t) = --83- t − --89- t2 + 2. a Find the height of the ball as it leaves the thrower’s hand. b Find when and where the ball reaches its greatest height. c Find when the ball returns to the same level that it left the thrower’s hand. d If the ball isn’t hit, find when the ball hits the ground to the nearest thousandth of a second. e Hence state the domain and range of h(t). f Sketch the graph of h versus t.

9C

2 A piece of wire of length 100 cm is to be cut so that one piece is used to form a square, while the other is used to form a circle. If the edge length of the square is x cm, find, in terms of x, a the radius of the circle b the area of the circle, and c the total area of the two shapes. Show that, when x = 14, the total area is minimum.

CHAPTER

test yourself

9

Chap 06 SM Page 211 Thursday, October 12, 2000 10:59 AM

Integral calculus

6 VCE coverage Area of study Units 3 & 4 • Calculus

In this chapter 6A Substitution where the derivative is present in the integrand 6B Linear substitution 6C Antiderivatives involving trigonometric identities 6D Antidifferentiation using partial fractions 6E Definite integrals 6F Applications of integration 6G Volumes of solids of revolution 6H Approximate evaluation of definite integrals and areas

Chap 06 SM Page 212 Thursday, October 12, 2000 10:59 AM

212

Specialist Mathematics

Integration techniques and applications You will have seen in your Maths Methods course and elsewhere that some functions can be antidifferentiated (integrated) using standard rules. These common results are shown in the table below where the function f(x) has an antiderivative F(x).

f(x)

F(x) ax n + 1

ax n

--------------- + c n+1

1 --x-

logekx + c

e kx

e kx ------- + c k

sin kx

– cos kx ------------------ + c k

cos kx

sin kx -------------- + c k

sec2kx

tan kx -------------- + c k

1 -------------------- , x ∈ (–a, a) a2 – x2

x Sin –1 --- + c a

–1 -------------------- , x ∈ (–a, a) a2 – x2

x Cos –1 --- + c a

a ---------------a2 + x2

x Tan –1 --- + c a

In this chapter you will learn how to find antiderivatives of more complex functions using various techniques.

Technique 1: Substitution where the derivative is present in the integrand d[ f ( x ) ]n + 1 Since ---------------------------- = ( n + 1 ) f ′( x ) [ f ( x ) ] n , n ≠ −1, as an application of the chain rule, dx then it follows that:

∫

[ f ( x ) ]n + 1 f ′( x ) [ f ( x ) ] n dx = ------------------------- + c, n ≠ – 1. n+1

d [ log e f ( x ) ] f ′( x ) Since ----------------------------= ------------ ; f(x) ≠ 0 dx f ( x) then it follows that

f ′( x )

- dx = log f ( x ) + c . ∫ ----------f ( x) e

The method relies on the derivative, or multiple of the derivative, being present and recognisable. Then, the appropriate substitutions may be made according to the above rules.

Chap 06 SM Page 213 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

213

WORKED Example 1 Find the antiderivative of the following expressions. 3 x2 + 1 a (x + 3)7 b 4x(2x2 + 1)4 c ------------------x3 + x THINK

WRITE

a

a Let u = x + 3.

b

1

Recognise that the derivative of x + 3 is 1. Let u = x + 3.

2

du Find ------ . dx

3

Make dx the subject.

or dx = du

4

Substitute for x + 3 and dx.

So (x + 3)7 dx = u7 du

5

Antidifferentiate with respect to u.

u8 = ----- + c 8

6

Replace u with x + 3 and state answer in terms of x.

( x + 3 )8 = ------------------- + c 8

1

Recognise that 4x is the derivative of 2x 2 + 1. Let u = 2x 2 + 1.

2

du Find ------ . dx

3

Make dx the subject.

4

du Substitute u for 2x 2 + 1 and ------ for dx. 4x

du ------ = 1 dx

∫

∫

b Let u = 2x2 + 1. du ------ = 4x dx dx or ------ = du 4x

∫

So 4x(2x2 + 1)4 dx du = 4x u4 -----4x

c

∫ = ∫u4 du

5

Simplify the integrand by cancelling out the 4x.

6

Antidifferentiate with respect to u.

u5 = ----- + c 5

7

Replace u with 2x 2 + 1.

( 2x 2 + 1 ) 5 = ------------------------- + c 5

1

2

Recognise that 3x 2 + 1 is the derivative of x3 + x. Let u = x3 + x. du Find ------ . dx

c Let u = x3 + x. du ------ = 3x 2 + 1 dx Continued over page

Chap 06 SM Page 214 Thursday, October 12, 2000 10:59 AM

214

Specialist Mathematics

THINK 3

4

WRITE du or dx = ----------------3x 2 + 1

Make dx the subject. du - for Substitute u for x 3 + x and ----------------3x 2 + 1 dx.

So

∫

3x 2 + 1 ------------------dx x3 + x

∫ du = ------∫u = u du ∫

=

3x 2 + 1 du ------------------ × ----------------3x 2 + 1 u

5

Cancel out 3x 2 + 1.

6

Express the integrand in index form.

7

Antidifferentiate with respect to u.

= 2u 2 + c

8

Replace u with x 3 + x.

= 2( x3 + x )2 + c

9

Express in root notation.

= 2 x3 + x + c

1 – --2

1 ---

1 ---

WORKED Example 2 Antidifferentiate the following functions with respect to x. x+3 a f ( x ) = ------------------------b f ( x ) = ( x 2 – 1 ) cos ( 3 x – x 3 ) ( x2 + 6 x )3 THINK WRITE a

1 2 3 4

Express in integral notation. Recognise that x + 3 is half of the derivative of x 2 + 6x. Let u = x 2 + 6x. du Find ------ . dx

5

Make dx the subject.

6

du Substitute u for x 2 + 6x and --------------- for dx. 2x + 6

7

Factorise 2x + 6.

8

Cancel out x + 3 and express u in index form on the numerator.

a

∫

x+3 ------------------------ dx 2 ( x + 6x ) 3

Let u = x2 + 6x. du ------ = 2x + 6 dx du or dx = --------------2x + 6 x+3 - dx = So -----------------------( x 2 + 6x ) 3

∫

- × --------------2x + 6 ∫ ----------u du x+3 = ------------ × -------------------2 ( x + 3) ∫u = u du ∫ x+3 3

3

1 –3 --2

du

Chap 06 SM Page 215 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

b

Antidifferentiate with respect to u.

10

Replace u with x2 + 6x.

11

Express the answer with a positive index number. (Optional.)

1

Express in integral notation.

3 4

215

WRITE

9

2

Integral calculus

u –2 = – ------- + c 4 ( x 2 + 6x ) –2 = – --------------------------- + c 4 1 = – ------------------------+c 2 4( x + x )2 b

Recognise that x2 − 1 is a multiple of the derivative of 3x − x3. Let u = 3x − x3. du Find ------ . dx

5

Make dx the subject.

6

du Substitute u for 3x − x3 and ----------------2– 3 3x for dx.

∫( x – 1 ) cos( 3x – x ) dx 2

3

Let u = 3x − x3. du ------ = 3 – 3x 2 dx du or dx = ----------------23 – 3x So

∫( x – 1 ) cos( 3x – x ) dx du = ( x – 1 ) cos u × ----------------∫ 3 – 3x du = ( x – 1 ) cos u × ---------------------∫ 3(1 – x ) du = ( x – 1 ) cos u × ------------------------∫ –3 ( x – 1 ) – cos u = ---------------- du ∫ 3 2

2

2

2

7

Factorise 3 − 3x2.

2

2

2

2

8

Cancel out x2 − 1.

9

Antidifferentiate with respect to u.

10

Replace u with 3x − x3.

– sin u = --------------- + c 3 – sin ( 3x – x 3 ) = -------------------------------- + c 3

WORKED Example 3

Evaluate the following indefinite integrals. x Tan –1 --b a cos x sin 4x dx 2 ------------------d x 4 + x2

∫

∫

c

THINK a 1 Recognise that cos x is the derivative of sin x. 2 Let u = sin x.

∫

log e 4 x ----------------d x x

d

∫sin 2x cos 3x dx

WRITE a Let u = sin x. Continued over page

Chap 06 SM Page 216 Thursday, October 12, 2000 10:59 AM

216

Specialist Mathematics

THINK

b

WRITE

3

du Find ------ . dx

4

Make dx the subject.

du or dx = -----------cos x

5

du Substitute u for sin x and ------------ for dx. cos x

So cos x sin4x dx =

6

Cancel out cos x.

7

Antidifferentiate with respect to u.

= 1--5- u5 + c

8

Replace u with sin x.

= 1--5- sin5x + c

1

du ------ = cos x dx

1 Recognise that -------------2- is half of the 4+x x derivative of Tan−1 --- . 2 x Let u = Tan−1 --- . 2

3

du Find ------ . dx

4

Make dx the subject.

b

x Let u = Tan−1 --- . 2 du 2 ------ = -------------2dx 4+x ( 4 + x 2 )du or dx = -------------------------2

( 4 + x 2 )du Substitute u for Tan --- and -------------------------2 2 for dx. −1 x

So

∫

x Tan –1 --2 ----------------2- dx 4+x ( 4 + x 2 )du u -------------2- × -------------------------2 4+x

∫ u = --- du ∫2 =

6

Cancel out 4 + x2.

7

Antidifferentiate with respect to u.

u2 = ----- + c 4

8

x Replace u with Tan−1 --- . 2

 Tan –1 --x-  2 = ------------------------- + c 4

2

c

1 2 3 4

4

4

2

5

du ∫ ( cos x )u ----------cos x = u du ∫

∫

1 Recognise that --- is the derivative of loge4x. x Let u = loge4x. du Find ------ . dx Make dx the subject.

c Let u = loge4x. du 1 ------ = --dx x or dx = x du

Chap 06 SM Page 217 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK 5

Substitute u for loge4x and x du for dx in the integral.

So

- dx ∫ --------------x log e 4x

∫ = u du ∫

u --- × x du x

6

Cancel out x.

7

Antidifferentiate with respect to u.

= 1--2- u2 + c

8

Replace u by loge4x.

= 1--2- (loge4x)2 + c

1

Express cos3x as cos x cos2x.

2 3

4

217

WRITE

=

d

Integral calculus

Express cos x cos2x as cos x (1 − sin2x) (using the identity sin2x + cos2x = 1). Let u = sin x as its derivative is a factor of the new form of the function. du Find ------ . dx

5

Make dx the subject.

6

du Substitute u for sin x and ------------ for dx. cos x

d

∫ sin2x cos3x dx = ∫ sin2x cos x cos2x dx = ∫ sin2x cos x (1 − sin2x) dx Let u = sin x. du ------ = cos x dx du or dx = -----------cos x So ∫ sin2x cos3x dx

7

Cancel out cos x.

du = ∫ u2 cos x (1 − u2) -----------cos x 2 2 = ∫ u (1 − u ) du

8

Expand the integrand.

= ∫ (u2 − u4) du

9

Antidifferentiate with respect to u.

= 1--3- u3 − 1--5- u5 + c

10

Replace u by sin x.

= 1--3- sin3x − 1--5- sin5x + c

WORKED Example 4

If f′(x) = 4xe x and f(0) = 5, find f(x). THINK 1 Express f(x) in integral notation. 2

2 3 4

Recognise that 4x is twice the derivative of x2. Let u = x2. du Find ------ . dx

WRITE f(x) = ∫ 4xe x dx 2

Let u = x2. du ------ = 2x dx

Continued over page

Chap 06 SM Page 218 Thursday, October 12, 2000 10:59 AM

218

Specialist Mathematics

THINK

WRITE

5

Make dx the subject.

du or dx = -----2x

6

du Substitute u for x2 and ------ for dx. 2x

So f(x) =

7

Cancel out 2x.

8

11

Antidifferentiate with respect to u. Replace u by x2. Substitute x = 0 and f(0) = 5. Solve for c.

12

State the function f(x).

9 10

∫ 4xe -----2x= 2e du ∫ u

du

u

= 2eu + c 2 f(x) = 2e x + c f(0) = 2e0 + c = 5 2+c=5 c=3 2 Therefore f(x) = 2e x + 3.

remember remember d[ f ( x ) ]n + 1 1. Since ---------------------------- = ( n + 1 ) f ′( x ) [ f ( x ) ] n, n ≠ – 1 dx [ f ( x ) ]n + 1 f ′( x ) [ f ( x ) ] n dx = ------------------------- + c, n ≠ – 1 . n+1 d log e f ( x ) f ′( x ) - = -----------2. Since ------------------------dx f ( x) f ′( x ) then ------------ dx = log e f ( x ) + c . f ( x) then

∫ ∫

6A Mat

d hca

WORKED

Example

Antidifferentiation

1&2

Substitution where the derivative is present in the integrand

1 Find the antiderivative for each of the following expressions. a 2x(x2 + 3)4

b 2x(6 − x2)−3

c

3x2(x3 − 2)5

d 2(x + 2)(x2 + 4x)−3

e

( 2x + 5 ) x 2 + 5x

f

2x – 3 -----------------------( x 2 – 3x ) 4

h

3x 2 + 4x -----------------------x 3 + 2x 2

j

(2x + 3) sin(x2 + 3x − 2)

l

cos x sin3x

g 3x2(x3 − 5)2 i

4x3e x

4

k (3x2 + 5) cos(x3 + 5x)

Chap 06 SM Page 219 Thursday, October 12, 2000 10:59 AM

Chapter 6

219

Integral calculus

log e x ------------x

m −sin4x cos x

n

o sec2x tan3x

( Sin –1 x ) 2 p ------------------------1 – x2

2 multiple choice Given that the derivative of (x2 + 5x)4 is 4(2x + 5)(x2 + 5x)3, then the antiderivative of 8(2x + 5)(x2 + 5x)3 is: A 2(x2 + 5x)4 + c

B

1 2 --- (x 2

+ 5x)4 + c

D 2(x2 + 5x)2 + c

E

1 2 --- (x 2

+ 5x)2 + c

C 4(x2 + 5x)4 + c

3 multiple choice dx can be found by making the substitution ‘u’ equal to: ∫ -----------------x +3 x

a The integral

2

A x2

B x

D x2 + 3

x

C

E 2x

b After the appropriate substitution the integral becomes: A D c

1 ---

∫ ∫ (u + 3) u 2 du

B 1 --2

du

2 2 --- ( x 3

3 ---

+ 3)2 + c 1 ---

D ( x2 + 6 )2 + c Example

2

u

1 – --2

du

∫

C

1 --2

C

2 2 --- ( x 3

(u + 3)

a 6x2(x3 − 2)5

1 ---

B 4( x2 + 3 )2 + c

x2(x3 − 1)7

1 ---

b x(4 − x2)3 d (x + 3)(x2 + 6x − 2)4 4x + 6 ---------------------x 2 + 3x

f

2x – 5 g --------------------------------( x 2 – 5x + 2 ) 6

h ( x2 – 1 )

4 – 3x + x 3

j

x2e x + 2

k (x + 1) sin(x2 + 2x − 3)

l

(x2 − 2) cos(6x − x3)

m sin 2x cos42x

n cos 3x sin23x

log e 3x o ---------------2x

(4x – 2) log e (x 2 – x) p ---------------------------------------------------x2 – x

2

(6x − 3)e x

−x+3

3 ---

+ 6)2 + c

E ( x2 + 3 )2 + c

e (x + 1)(x2 + 2x + 3)−4

i

du

du

4 Antidifferentiate each of the following expressions with respect to x. c

1 – --2

x - is: Hence the antiderivative of ------------x2 + 3 A

WORKED

E

1 – --2

∫ 2 u ∫

1 --2

Chap 06 SM Page 220 Thursday, October 12, 2000 10:59 AM

220 WORKED

Example

3

Specialist Mathematics

5 Evaluate the following indefinite integrals. a c e g

i k m o

5 ---

∫ ∫ e ( 3 + 2e ) dx ∫ x sin x dx cos x log ( sin x ) - dx ∫ -----------------------------------------sin x x ( x 2 + 1 ) 2 dx x 4

x

2

3

e

∫

x – 2 Cos –1 --3 --------------------------- dx 9 – x2

∫ Sin 4x dx ∫ -----------------------1 – 16x x - dx ∫ 1---------------– 4x

( x + 2 ) cos ( x 2 + 4x ) dx –1

2

b d f h

j

l n

∫ x 1 – x dx sin x - dx ∫ -----------cos x ∫ sin x e dx ∫ e ( 1 – e ) dx 2

3

cos x

3x

3x 2

∫ ( 2x + 1 )

x + x 2 – 3 dx

∫

e x+1 ---------------- dx x+1

∫

Tan –1 x ------------------ dx 1 + x2

2

6 Find the antiderivative for each of the following expressions. cos x a -----------------------------b sec 2 x 2 + tan x 1 + 3 sin x c

sin x sec3x

e

sec 2 x ---------------------------3( 5 – tan x )

( log e x ) 3 g -------------------x e x – e–x ---------------------e x + e–x k sin3x cos2x log e ( tan x ) m --------------------------sin x cos x i

WORKED

Example

4

e2 x d ---------------------( e2 x – 3 )2 4 f ---------------xlog e x h

e tan x -----------cos 2 x

j

sin x – cos x -----------------------------sin x + cos x

l

cos3x sin4x

x 7 If f ′( x ) = ------------------ and f(2) = 1 find f(x). 2 x +5 e x 8 If f ′( x ) = -------- and f(0) = 3 find f(x). x 4 log e x 2 - then find g(x). 9 If g(1) = −2 and g′ ( x ) = ------------------x

π 10 If g  --- = 0 and g′(x) = 16 sin x cos3x then find g(x).  4

Chap 06 SM Page 221 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

221

Technique 2: Linear substitution For antiderivatives of the form

∫ f ( x)[g( x)]

n

dx, n ≠ 0 where g(x) is a linear function,

that is, of the type mx + c, and f(x) is not the derivative of g(x), the substitution u = g(x) is often successful in finding the integral. Examples of this type of integral are: 1.

∫

4x + 1 dx . In this example f(x) = 1 and g(x) = 4x + 1 with n =

u = 4x + 1, and consequently dx = 1--4- du, the integral becomes readily antidifferentiated. 2.

∫ 4x( x – 3 )

4

1 --4

∫

1 --- . 2

By letting

u du which can be

dx . In this example f(x) = 4x and g(x) = x − 3 with n = 4. By letting

u = x − 3, the function f(x) can be written in terms of u, that is, u = x − 3, thus 4x = 4(u + 3) and further, dx = du. The integral becomes

∫ 4( u + 3 ) × u du which

can be readily antidifferentiated. The worked examples below illustrate how the use of the substitution u = g(x) simplifies integrals of the type

∫ f ( x)[g( x)]

n

dx .

WORKED Example 5

i Using the appropriate substitution, express the following integrals in terms of u only. ii Evaluate the integrals as functions of x. a

∫

5 ---

x( x – 2)2 d x

b

∫

x2 ---------------- d x x+1

THINK a i

1 2 3 4

ii

WRITE Let u = x − 2. du Find ------ . dx Make dx the subject. Substitute u for x − 2, u + 2 for x and du for dx.

5

Expand the integrand.

1

Antidifferentiate with respect to u.

2

Replace u with x − 2.

a i Let u = x − 2 and x = u + 2. du ------ = 1 dx dx = du 5 ---

∫ = ( u + 2 )u du ∫ =  u + 2u  dx  ∫ So

x ( x – 2 ) 2 dx 5 --2

7 --2

9 ---

5 --2

7 ---

ii = 2--9- u 2 + 4--7- u 2 + c 9 ---

7 ---

= 2--9- ( x – 2 ) 2 + 4--7- ( x – 2 ) 2 + c Continued over page

Chap 06 SM Page 222 Thursday, October 12, 2000 10:59 AM

222

Specialist Mathematics

THINK 3

WRITE Take out the factor of

7

--- x – 2 4 = 2 ( x – 2 ) 2  ----------- + --- + c  9 7

7 --2

2( x – 2) . 4

Simplify the other factor.

7 --- 7x – 14 + 36 = 2 ( x – 2 ) 2  ------------------------------ + c   63 7 --- 7x + 22 = 2 ( x – 2 ) 2  ------------------ + c  63 

b i

1

Express x + 1 in index form.

b i

∫

x2 ---------------- dx x+1

∫

= 2

Let u = x + 1.

3

du Find ------ . dx

4

Make dx the subject.

5

Express x in terms of u.

6

Hence express x2 in terms of u.

7

ii

x2( x + 1 )

1 2

– ---

dx

Let u = x + 1. du ------ = 1 dx

Substitute u for x + 1, u2 − 2u + 1 for x2 and du for dx.

8

Expand the integrand.

1

Antidifferentiate with respect to u.

2

Replace u with x + 1.

3

Take 2 ( x + 1 ) 2 out as a factor.

4

Simplify the other factor.

1 ---

or dx = du x=u−1 x2 = u2 − 2u + 1 So

∫

x ( x + 1 ) – 2 dx

=

∫

( u 2 – 2u + 1 )u

=

∫

 u --2- – 2u --2- + u – --2- du  

1 ---

2

1

3

5 ---

3 ---

1 2

– ---

du

1

1 ---

ii = 2--5- u 2 – 4--3- u 2 + 2u 2 + c 5 ---

3 ---

1 ---

= 2--5- ( x + 1 ) 2 – 4--3- ( x + 1 ) 2 + 2 ( x + 1 ) 2 + c 1 --- ( x + 1 ) 2 2( x + 1) = 2 ( x + 1 ) 2 ------------------- – -------------------- + 1 + c 5 3 1 --- ( x 2 + 2x + 1 ) ( – 2x – 2 ) = 2 ( x + 1 ) 2 -------------------------------- + ------------------------ + 1 + c 5 3 1 --- 3x 2 + 6x + 3 – 10x – 10 + 15 = 2 ( x + 1 ) 2 ---------------------------------------------------------------------- + c 15 1 --- 3x 2 – 4x + 8 = 2 ( x + 1 ) 2 ------------------------------ + c 15

Chap 06 SM Page 223 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

223

WORKED Example 6 e2 x -. a Find the antiderivative of ------------ex + 1

b State the domain of the antiderivative.

THINK a

1

WRITE a Let u = e x + 1.

Since e2x = (e x)2, it can be antidifferentiated the same as a linear function by letting u = e x + 1.

2

du Find ------ . dx

du ------ = e x dx

3

Make dx the subject.

4

Express e x in terms of u.

and e x = u − 1

5

du Substitute u for e + 1 and -----x- for dx. e

So

du or dx = -----xe

x

∫

e2 x ------------- dx x e +1 e x e x du ---------- × -----xu e

∫ e = ----- du ∫u u–1 = ------------ du ∫ u 1 =  1 – --- du ∫  u

=

b

x

6

Cancel out e x.

7

Substitute u − 1 for the remaining e x.

8

Simplify the rational expression.

9

Antidifferentiate with respect to u.

= u − logeu + c

10

Replace u with e x + 1.

= e x + 1 − loge(e x + 1) + c

1

2

e x + 1 > 0 for all values of x as e x > 0 for all x. The function loge f(x) exists wherever f(x) > 0. State the domain.

b For loge(e x + 1) to exist e x + 1 > 0, which it is for all x. Therefore the domain of the integral is R.

Note: Recall that the logarithm of a negative number cannot be found.

remember remember For antiderivatives of the form

∫ f ( x)[g( x)]

n

dx, n ≠ 0 , make the substitution

u = g(x) and so [g(x)]n dx, n ≠ 0 becomes g′(x) un du, n ≠ 0. This technique can be used for the specific case where g = mx + c since g′(x) = m. The function f(x) needs to be transformed in terms of the variable u as well.

Chap 06 SM Page 224 Thursday, October 12, 2000 10:59 AM

224

Specialist Mathematics

6B Mat

d hca

Linear substitution

1 By making the appropriate substitution for u: i express the following integrals in terms of u 5 ii evaluate the integrals as functions of x. 4 2 ----------- dx --------------- dx a b x–3 3x + 5

WORKED

Example

Antidifferentiation

c e g i k m o

∫ ∫ 4x + 1 dx ∫ x( x + 1 ) dx ∫ 2x( 2x + 1 ) dx ∫ 6x( 3x – 2 ) dx ∫ x x + 3 dx ∫ ( x + 2 )( x – 4 ) dx 2x - dx ∫ --------------x–6

∫ ∫ 3 – 2x dx ∫ 4x( x – 3 ) dx ∫ 3x( 1 – 3x ) dx ∫ x( 2x + 7 ) dx ∫ x 3x – 4 dx ∫ ( x – 3 )( 2x + 1 ) 3x - dx ∫ --------------8–x

d f

3

4

h

4

3 --4

5

1 --3

j l

3 --2

n p

5 --2

dx

2 multiple choice a The integral

∫ 4x

x+2

A

x + 2 dx can be found by letting u equal: C x+2

B x

D 4x

E 2x

b The integral then becomes: 5 ---

A

∫

u 2 du

D

∫

 4u --2- – 2u --2- du  

1

1

1

B

∫

 2u --2- – 4u – --2- du  

E

∫

 4u --2- – 8u --2- du  

3

3

C

1 ---

∫

2u 2 du

∫

 u --2- + 2u – --2- du  

1

3 multiple choice

∫

a Using the appropriate substitution, A

∫

u du

D

∫

 u --2- + 2u --2- + u --2- du  

5

B 3

1

E

∫ ∫u

x2 ---------------- dx becomes: x–1

3

1

1

 u --2- + 2u --2- + u – --2- du   3 --2

du

C

1

1

Chap 06 SM Page 225 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

225

b The result of the integration is:

WORKED

Example

6a

Example

6b

2 --- ( x 3

– 1)2 + c

C

2 --- ( x 5

– 1)2 + c

E

2 --- ( x 7

– 1 ) 2 + 4--5- ( x – 1 ) 2 + 4--3- ( x – 1 ) 2 + c

5 ---

7 ---

5 ---

3 ---

1 ---

5 ---

3 ---

B

2 --- ( x 3

– 1)2 + 4( x – 1)2 + c

D

2 --- ( x 5

– 1 ) 2 + 4--3- ( x – 1 ) 2 + 2 ( x – 1 ) 2 + c

1 ---

3 ---

4 Find the antiderivative of each of the following expressions. a x2(x − 4)4

b x2(5 − x)3

d x2 3 – x

e

g ( x + 1 )2 x – 2

2

c

x

x2( x + 2 )3

f

x2( 1 – x )4

h ( x – 3 )2 x + 1

i

ex ------------x e +1 x3 x – 1

4 ---

x–1 3 ---

x2 ---------------x+1

k

2x 2 ---------------3–x

l

x3 m ---------------x+4

n

2x 3 ---------------1–x

x+3 o ------------------2( x – 2)

2x – 1 p ------------------3( x + 1)

4x q ------------------2( x + 2)

r

x2 ------------------2( x – 1)

( x – 2 )2 ------------------2–x

u

e2 x ------------x e +2

j

WORKED

3 ---

A

s

( x + 3 )2 ------------------x+2

v

e3 x ------------x e –1

t

1 ---

5 a If f ′( x ) = – ( 5 – x ) 2 + 10 ( 5 – x )

1 2

– ---

and f(1) = −2, find f(x).

b State the domain of f(x). 3 ---

1

– ---

1 --5( x + 1)2 ( x + 1) 2 6 a If f ′( x ) = ---------------------- – 3 ( x + 1 ) 2 + --------------------- and f(0) = 1, find f(x). 2 2

b State the domain of f(x). 2x + 1 7 a Given that g′ ( x ) = ------------------2- and g(2) = 0, find g(x). ( x – 1) b State the domain of g(x). e2 x - , find g(x). 8 a Given that g(0) = 2 – loge 2 and g′ ( x ) = ------------ex + 1 b State the domain of g(x).

Chap 06 SM Page 226 Thursday, October 12, 2000 10:59 AM

226

Specialist Mathematics

Technique 3: Antiderivatives involving trigonometric identities Different trigonometric identities can be used to antidifferentiate sinnx and cosnx; n ∈ J + depending on whether n is even or odd. Functions involving tan2ax are also discussed.

Even powers of sin x or cos x The double-angle trigonometric identities can be used to antidifferentiate even powers of sin x or cos x. The first identity is: cos 2x = 1 − 2 sin2x = 2 cos2x − 1 Therefore

sin2x = 1--2- (1 − cos 2x)

or

cos2x = 1--2- (1 + cos 2x)

The second identity is:

sin 2x = 2 sin x cos x sin x cos x =

or

1 --2

sin 2x

These may be expressed in the following general forms: sin2ax = 1--2- (1 – cos 2ax)

Identity 1

cos2ax = 1--2- (1 + cos 2ax)

Identity 2

sin ax cos ax =

Identity 3

1 --2

sin 2ax

WORKED Example 7 Find the antiderivative of the following expressions. x x a sin2 --b 2cos2 --2 4 THINK

WRITE

a

a

b

∫ sin --2- dx = ∫ ( 1 – cos x ) dx = ∫ ( 1 – cos x ) dx 2x

1

Express in integral notation.

2

x Use identity 1 to change sin2 --- . 2

3

Take the factor of

4

Antidifferentiate by rule.

= 1--2- ( x – sin x ) + c

5

Simplify the answer.

x 1 = --- – --- sin x + c 2 2

1

Express in integral notation.

1 --2

1 --2

1 --2

to the front of the integral.

b

∫ 2cos --4- dx 2x

Chap 06 SM Page 227 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

Integral calculus

227

WRITE

2

x Use identity 2 to change cos2 --- . 4

3

Simplify the integral.

4

Antidifferentiate by rule.

∫ x =  1 + cos --- dx  ∫ 2

x 2  1--2-  1 + cos --- dx  2

=

x = x + 2sin --- + c 2

WORKED Example 8

Evaluate the following indefinite integrals as functions of x. x x sin x cos x d x a b 4 sin 2 --- cos 2 --- d x 2 2 THINK WRITE

∫

a

b

1

∫

Use identity 3 to change sin x cos x. Note: The integral could be antidifferentiated using technique 1 since the derivative of sin x is cos x.

2

Antidifferentiate by rule.

1

x x Express sin2 --- cos2 --- as a perfect square. 2 2

a

∫ sin x cos x dx = ∫ sin 2x dx 1 --2

= – 1--4- cos 2x + c b

∫

x x 4 sin 2 --- cos 2 --- dx 2 2 x x 2 4  sin --- cos --- dx  2 2

∫ = 4 ( sin x ) dx ∫ = 4 ( sin x ) dx ∫ = sin x dx ∫ = ∫ ( 1 – cos 2x ) dx =

2

x x Use identity 3 to change sin --- cos --- . 2 2

3

Square the identity.

4

Simplify the integral.

5

Use identity 1 to change sin2x.

6

Antidifferentiate by rule.

= 1--2- ( x – 1--2- sin 2x ) + c

7

Simplify the answer.

x = --- – 1--4- sin 2x + c 2

1 --2 1 --4

2

2

2

1 --2

Odd powers of sin x or cos x For integrals involving odd powers of sin x or cos x the identity: sin2x + cos2x = 1 can be used so that the ‘derivative method’ of substitution then becomes applicable. The following worked example illustrates the use of this identity whenever there is an odd-powered trigonometric function in the integrand.

Chap 06 SM Page 228 Thursday, October 12, 2000 10:59 AM

228

Specialist Mathematics

WORKED Example 9

Find the antiderivative of the following expressions. a cos 3x b cos x sin 2x c cos 42x sin32x THINK

WRITE

a

a

1

Express in integral notation.

2

Factorise cos3x as cos x cos2x.

3

Use the identity: (1 − sin2x) for cos2x.

4

∫ cos x dx = cos x cos x dx ∫ = cos x ( 1 – sin x ) dx ∫ 3

2

2

Let u = sin x.

Let u = sin x so the derivative method can be applied.

5

du Find ------ . dx

du ------ = cos x dx

6

Make dx the subject.

du or dx = -----------cos x

7

du Substitute u for sin x and ------------ for dx. cos x

So

∫ cos x( 1 – sin x ) dx du = cos x ( 1 – u ) -----------∫ cos x = ( 1 – u ) du ∫ 2

2

b

2

8

Cancel out cos x.

9

Antidifferentiate with respect to u.

= u − 1--3- u3 + c

10

Replace u with sin x.

= sin x − 1--3- sin3x + c

1

Express in integral notation.

2

Use identity 3 in reverse to express sin 2x as 2 sin x cos x.

3

Simplify the integrand.

4

Let u = cos x so that the derivative method can be applied.

5

du Find ------ . dx

6

Make dx the subject.

7

du Substitute u for cos x and -------------- for dx. – sin x

b

∫ cos x sin 2x dx = cos x ( 2 sin x cos x ) dx ∫ = 2sin x cos x dx ∫ 2

Let u = cos x.

du ------ = – sin x dx du or dx = -------------– sin x

∫ 2 sin x cos x dx du = 2 sin x ( u ) -------------∫ – sin x 2

So

2

Chap 06 SM Page 229 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

c

Integral calculus

WRITE

∫ –2u

8

Cancel out sin x.

=

9

Antidifferentiate with respect to u.

= − 2--3- u3 + c

10

Replace u with cos x.

= − 2--3- cos3x + c

1

Express in integral notation.

2

Factorise sin32x as sin 2x sin22x.

=

∫ cos 2x sin 2x sin 2x dx

3

Use the identity 1 − cos22x for sin22x.

=

∫ cos 2x sin 2x( 1 – cos 2x ) dx

4

Let u = cos 2x so that the derivative method can be applied.

c

2

du

∫ cos 2x sin 2x dx 4

3

4

2

4

2

Let u = cos 2x.

5

du Find ------ . dx

6

Make dx the subject.

du or dx = ---------------------– 2 sin 2x

du Substitute u for cos 2x and ---------------------– 2 sin 2x for dx.

So

8

Cancel out sin 2x.

=

∫ – u ( 1 – u ) du

9

Expand the integrand.

=

∫ (u

10

Antidifferentiate with respect to u.

= 1--2- ( 1--7- u7 − 1--5- u5) + c

11

Simplify the result.

=

1 7 ------ u 14

12

Replace u with cos 2x.

=

1 7 ------ cos 2x 14

7

229

du ------ = – 2 sin 2x dx

∫u

4

du sin 2x ( 1 – u 2 ) ---------------------– 2sin 2x

1 4 --2

1 --2

6

−

2

– u 4 ) du

1 5 ------ u 10

Using the identity sec2x = 1 + tan2x

−

+c 1 5 ------ cos 2x 10

+c

The identity sec2ax = 1 + tan2ax is used to antidifferentiate expressions involving tan2ax + c where c is a constant since the antiderivative of sec2x is tan x. Otherwise, expressions of the form tannx sec2x can be antidifferentiated using the ‘derivative method’ of exercise 6A.

Chap 06 SM Page 230 Thursday, October 12, 2000 10:59 AM

230

Specialist Mathematics

WORKED Example 10

Find an antiderivative for each of the following expressions.

∫

a ( 2 + tan 2 x ) d x

∫

b 3 tan 2 3 x sec 2 3 x d x

THINK a

1

WRITE

Express 2 + tan2x as 1 + sec2x using the identity.

a

∫ ( 2 + tan x ) dx = ( 1 + sec x ) dx ∫ 2

2

2

b

1

Antidifferentiate by rule. There is no need to add c as one antiderivative only is required.

= x + tan x

Let u = tan 3x so that the derivative method can be applied.

b Let u = tan 3x.

2

du Find ------ . dx

du ------ = 3 sec 2 3x dx

3

Make dx the subject.

du or dx = --------------------3 sec 2 3x

4

du Substitute u for tan 3x and -------------------for dx. 3 sec 2 3x

So

∫ 3 tan 3x sec 3x dx du = 3 u sec 3x -------------------∫ 3sec 3x 2

2

2

2

2

∫u

5

Cancel out 3sec23x.

=

6

Antidifferentiate with respect to u.

= 1--3- u3

7

Replace u with tan 3x.

= 1--3- tan33x

2

du

remember remember 1. Trigonometric identities can be used to antidifferentiate odd and even powers of sin x and cos x. These identities are: sin2ax = 1--2- (1 − cos 2ax) cos2ax = 1--2- (1 + cos 2ax) sin ax cos ax =

1 --2

sin 2ax

2. The identity sec2ax = 1 + tan2ax is used to antidifferentiate expressions involving tan2ax + c where c is a constant.

Chap 06 SM Page 231 Thursday, October 12, 2000 10:59 AM

Chapter 6

6C WORKED

7

WORKED

Example

8

231

Antiderivatives involving trigonometric identities

1 Antidifferentiate each of the following expressions with respect to x. b sin22x c 2 cos24x d 4 sin23x a cos2x x x h sin2 --g cos2 --e cos25x f sin26x 2 3 2x 3x x x i 3 cos2 --j 2 sin2 --k cos2 -----l sin2 -----3 2 6 4 2 Evaluate the following indefinite integrals as functions of x. a c e g i k

∫ 2 sin x cos dx ∫ sin 3x cos 3x dx ∫ sin x cos x dx ∫ 2 sin 4x cos 4x dx x x 6 sin --- cos --- dx ∫ 2 2 5x 5x ∫ sin -----2- cos -----2- dx 2

d f

2

2

2

2

b

h

2

2

j

2

l

∫ 4 sin 2x cos 2x dx ∫ –2 sin 4x cos 4x dx ∫ sin 2x cos 2x dx ∫ 2 sin 3x cos 3x dx x x 4 sin --- cos --- dx ∫ 3 3 4x 4x ∫ –2 sin -----3- cos -----3- dx 2

2

2

2

2

2

2

2

3 multiple choice If a is a constant, then, a

b

∫ sin ax dx is equal to: 2

A 2x − sin 2ax + c

B x − 2asin 2ax + c

ax D x – 1--2- sin ------ + c 2

x 1 ax E --- – --- sin ------ + c 2 a 2

∫ sin ax cos ax dx is equal to: 2

2

x sin 4ax A --- – ------------------ + c 8 32a sin 4ax D x – ------------------ + c 16a c

x sin 2ax C --- – ------------------ + c 2 4a

∫ cos ax dx is equal to:

x sin ax B --- – -------------- + c 2 4a x cos 4ax E --- + ------------------- + c 8 16a

x cos ax C --- – --------------- + c 4 8a

3

A a cos ax − 3a cos3ax + c cos 4 ax C ---------------- + c 4a 1 E ------ ( 3 sin ax – sin 3 ax ) + c 3a

B a sin ax − 3 cos3ax + c sin 4 ax D --------------- + c 4a

Math

cad

Example

Integral calculus

Antidifferentiation

Chap 06 SM Page 232 Thursday, October 12, 2000 10:59 AM

232 WORKED

Example

9a

WORKED

Example

9b

Specialist Mathematics

4 Find an antiderivative of each of the following expressions. a sin3x b cos32x c 6 sin34x x g 3 sin 3 --e sin37x f cos36x 2 3x 5x 3x i sin 3 -----j cos 3 -----k sin 3 -----2 2 4

d 4 cos3 3x x h 2 cos 3 --3 4x l cos 3 -----3

5 Use the appropriate identities to antidifferentiate the following expressions. a sin x cos 2x b cos 2x cos 4x c sin 3x cos 6x x x 2x d cos 4x cos 8x e sin --- cos x f cos --- cos -----2 3 3 6 Antidifferentiate each of the following expressions with respect to x. x x c sin --- cos 5 --b sin 2x cos32x a sin x cos4x 2 2 x 2x 2x x e cos --- sin 6 --f cos ------ sin 7 -----d cos 3x sin43x 5 3 3 5

WORKED

Example

9c

7 Find the following integrals. a c e g i k m

WORKED

Example

10

∫ cos x sin x dx ∫ cos 2x sin 2x dx x x cos --- sin --- dx ∫ 2 2 x x ∫ 4 cos --3- sin --3- dx ∫ sin x cos dx ∫ 2sin 2x cos 2x dx x x 4 sin --- cos --- dx ∫ 2 2 2

b

3

2

d

3

2

f

3

2

3

h

3

4

3

3

j

5

l

6

n

∫ sin x cos x dx ∫ sin 3x cos 3x dx 3x 3x sin ------ cos ------ dx ∫ 2 2 5x 5x ∫ –6 sin -----4- cos -----4- dx ∫ cos 2x sin 2x dx ∫ –2 cos 3x sin 3x dx 3x 3x cos ------ sin ------ dx ∫ 2 2 2

3

2

3

2

3

2

3

3

4

3

6

3

7

8 Find an antiderivative for each of the following expressions: x b 1 + tan 2 --c tan2x sec2x a 1 + tan22x 3 x x f 8 tan 4 --- sec 2 --e 4 tan52x sec22x d tan3x sec2x 2 2 x x i 2 tan 2 --- sec 4 --h 6 tan22x sec42x g tan2x sec4x 2 2 x x k tan 4 --- sec 4 --l 12 tan56x sec66x j 3 tan33x sec43x 5 5 9 Find the following integrals where n ∈ J+. a d

∫ sin x cos x dx ∫ sin x cos x dx 3

n

b

n

e

∫ cos x sin x dx ∫ cos x sin x dx n

3

n

c

∫ sec x tan x dx 2

n

Chap 06 SM Page 233 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

233

π 10 If f ′(x) = 6 sin x cos2x and f  --- = 0 , find f(x).  3 π 11 If f ′(x) = 4 sin22x cos22x and f  --- = π , find f(x).  4 x x 4 12 Find g(x) if g′(x) = sin3 --- cos4 --- and g(0) = – ------ . 2 2 35

The graph of a function and the graphs of its antiderivatives 2x Given f(x) = 2cos --- , what do the graphs of its 2 antiderivatives look like?

Using a graphics calculator, press Y= , enter Y1= 2(cos(X/2))2, move down to Y2 = , press MATH and select 9:fnInt(. Complete to obtain fnInt(Y1,X,0,X).

(Remember that to insert the symbol Y1, press VARS , select Y-VARS and 1:Function. Then select 1:Y1 and press ENTER (and similarly for any Y variable).

As the given function is trigonometric, press ZOOM and select 7: ZTrig.

(Since the numeric integral is repeatedly applied for every X-value on the screen, the antiderivative graph can take some time to plot. You can speed it up considerably by changing the value of Xres in the WINDOW settings to 5.) 2x 1 Which is the graph of f(x) = 2cos --- and which is the graph of the antiderivative? 2

The antiderivative graph in the second screen is the line that cuts 0 at x = 0, since the integral from 0 to 0 of any function is 0. To see another antiderivative graph, go to Y3 = , press MATH , select 9 and complete 9: fnInt(Y1,X,1,X) and then press GRAPH . 2 Generate another two antiderivative graphs on your calculator. Sketch the function and the four antiderivative graphs. Describe any relationships you can find. 3 Choose another function and investigate the relationship between the graph of the function and the graphs of its antiderivatives.

Chap 06 SM Page 234 Thursday, October 12, 2000 10:59 AM

234

Specialist Mathematics

Technique 4: Antidifferentiation using partial fractions Recall that rational expressions, in particular those with denominators that can be expressed with linear factors, can be transformed into partial fractions. A summary of two common transformations is shown in the table below. These transformations are useful when the degree of the numerator is less than the degree of the denominator; otherwise long division is generally required before antidifferentiation can be performed.

Rational expression

Equivalent partial fraction

f ( x) ---------------------------------------( ax + b ) ( cx + d ) where f(x) is a linear function

A B -------------------- + -------------------( ax + b ) ( cx + d )

f ( x) ---------------------( ax + b ) 2 where f(x) is a linear function

A B ---------------------- + -------------------( ax + b ) 2 ( ax + b )

We have seen how this procedure simplifies the sketching of graphs of rational functions. Similarly, expressing rational functions as partial fractions enables them to be antidifferentiated quite easily. However, it is preferable to use a substitution method, if it is applicable, as the partial-fraction technique can be tedious.

WORKED Example 11

Find a, b and c if ax(x − 2) + bx(x + 1) + c(x + 1)(x − 2) = 2x − 4. THINK WRITE Let x = 0, −2c = −4 1 Let x = 0 so that c can be evaluated. c=2 2 Solve the equation for c. Let x = 2, 6b = 0 3 Let x = 2 so that b can be evaluated. b=0 4 Solve the equation for b. Let x = −1, 3a = −6 5 Let x = −1 so that a can be evaluated. a = −2 6 Solve the equation for a. State the solution. Therefore a = −2, b = 0 and c = 2. 7

WORKED Example 12

For each of the following rational expressions: i express as partial fractions ii antidifferentiate the result. x+7 2x – 3 a ----------------------------------b -------------------------( x + 2)( x – 3) x2 – 3 x – 4 THINK a i 1 Express the rational expression as two separate fractions with denominators (x + 2) and (x − 3) respectively. 2 Express the partial fractions with the original common denominator.

WRITE x+7 a b a i ---------------------------------- = ----------------- + ---------------( x + 2)( x – 3) ( x + 2) ( x – 3) a( x – 3) + b( x + 2) = ---------------------------------------------( x + 2)( x – 3)

Chap 06 SM Page 235 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

b i

ii

235

WRITE x + 7 = a(x − 3) + b(x + 2)

so

4

Equate the numerator on the lefthand side with the right-hand side. Let x = −2 so that a can be evaluated.

5

Solve for a.

6

Let x = 3 so that b can be evaluated.

7

Solve for b.

8

Rewrite the rational expression as partial fractions.

x+7 –1 2 Therefore ---------------------------------- = ------------ + ----------( x + 2)( x – 3) x+2 x–3

1

Express the integral in partial fraction form.

ii

2 3

Antidifferentiate by rule. Simplify using log laws.

1

Factorise the denominator.

2

Express the partial fractions with denominators (x − 4) and (x + 1) respectively.

3 4

Express the right-hand side with the original common denominator. Equate the numerators.

So

5

Let x = 4 to evaluate a.

Let x = 4,

6

Solve for a.

7

Let x = −1 to evaluate b.

8

Solve for b.

9

Rewrite the rational expression as partial fractions.

1

Express the integral in its partial fraction form.

2

Antidifferentiate by rule. Simplify using log laws.

3

ii

Integral calculus

3

Let x = −2, and thus 5 = −5a a = −1 Let x = 3, and thus 10 = 5b b=2

- dx ∫ --------------------------------( x + 2)( x – 3) –1 2 =  ------------ + ----------- dx ∫  x + 2 x – 3 x+7

= −loge(x + 2) + 2 loge(x − 3) + c (x > 3) 2

( x – 3) = loge ------------------- + c x+2 2x – 3 2x – 3 - = ---------------------------------b i ------------------------2 ( x – 4)( x + 1) x – 3x – 4 a b = ----------- + -----------x–4 x+1 a( x + 1) + b( x – 4) = ---------------------------------------------( x – 4)( x + 1) 2x − 3 = a(x + 1) + b(x − 4) 5 = 5a a=1 Let x = −1, −5 = −5b b=1 2x – 3 1 1 - = ----------- + -----------Therefore ------------------------2 x – 3x – 4 x – 4 x + 1 ii

- dx ∫ ------------------------x – 3x – 4 1 1 =  ----------- + ------------ dx ∫  x – 4 x + 1 2x – 3

2

= loge(x − 4) + loge(x + 1) + c (x > 4) = loge[(x − 4)(x + 1)] + c (x > 4)

or loge(x2 − 3x − 4) + c (x > 4)

Chap 06 SM Page 236 Thursday, October 12, 2000 10:59 AM

236

Specialist Mathematics

WORKED Example 13 Find the following integrals. x2 + 6 x – 1 2 -------------2- d x a b ----------------------------------- d x ( x + 4)( x + 1) 1–x

∫

∫

THINK a

WRITE 2 2 -------------- = ---------------------------------1 – x2 ( 1 – x )( 1 + x )

1

Factorise the denominator of the integrand. a

2

Express into partial fractions with denominators (1 − x) and (1 + x).

3

Express the partial fractions with the original common denominator.

4

Equate the numerators.

so

5

Let x = 1 to find a.

Let x = 1, 2 = 2a

6

Solve for a.

7

Let x = −1 to find b.

8

Solve for b.

9

Express the integrand in its partial fraction form.

a b = ----------- + -----------1–x 1+x a(1 + x) + b(1 – x) = ---------------------------------------------1 – x2

a=1 Let x = −1, 2 = 2b b=1 Therefore =

b

10

Antidifferentiate by rule.

11

Simplify using log laws.

1

The degree of the numerator is the same as the degree of the denominator and hence the denominator should divide the numerator using long division.

2

Expand the denominator.

3

Divide the denominator into the numerator.

2 = a(1 + x) + b(1 − x)

- dx ∫ 1------------–x 2

2

+ ------------ dx ∫  ----------1 – x 1 + x 1

1

= −loge(1 − x) + loge(1 + x) + c, (−1 < x < 1) 1+x = loge  ------------ + c  1 – x

(−1 < x < 1)

b

x 2 + 6x – 1 x 2 + 6x – 1 ---------------------------------- = -------------------------( x + 4)( x + 1) x 2 + 5x + 4 Using long division: 1 2 2 x + 5x + 4 ) x + 6x − 1 x2 + 5x + 4 x−5 The division yields 1 with remainder (x − 5).

Chap 06 SM Page 237 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK 4

5

Integral calculus

237

WRITE

Rewrite the rational expression using the result of the division.

Therefore x 2 + 6x – 1 x–5 ---------------------------------- = 1 + ---------------------------------( x + 4)( x + 1) ( x + 4)( x + 1)

x–5 Express ---------------------------------- as partial ( x + 4)( x + 1) fractions with denominators (x + 4) and (x + 1).

x–5 a b Now ---------------------------------- = ------------ + -----------( x + 4)( x + 1) x+4 x+1

a( x + 4) + b( x + 1) = ----------------------------------------------( x + 4)( x + 1)

6

Rewrite the partial fractions with the original common denominator.

7

Equate the numerators.

and thus x − 5 = a(x + 4) + b(x + 1)

8

Let x = −1 to find a.

Let x = −1, −6 = 3a

9

Solve for a.

10

Let x = −4 to find b.

11

Solve for b.

12

Express the original integrand in its partial fraction form.

a = −2 Let x = −4, −9 = −3b b=3 Therefore, =

13

Antidifferentiate by rule.

∫

x 2 + 6x – 1 ---------------------------------- dx ( x + 4)( x + 1)

- + ------------ dx ∫  1 + ----------x + 4 x + 1 –2

3

= x − 2loge(x + 4) + 3loge(x + 1) + c, (x > −1).

remember remember Rational polynomials can be antidifferentiated by rewriting the expressions as partial fractions or by long division. If the numerator is of degree less than the denominator then use partial fractions; otherwise rewrite the expression by long division. Two common partial fraction transformations are shown below.

Rational expression

Equivalent partial fraction

f ( x) ---------------------------------------( ax + b ) ( cx + d ) where f(x) is a linear function

A B -------------------- + -------------------( ax + b ) ( cx + d )

f ( x) ---------------------( ax + b ) 2 where f(x) is a linear function

A B ---------------------- + -------------------( ax + b ) 2 ( ax + b )

Chap 06 SM Page 238 Thursday, October 12, 2000 10:59 AM

238

Specialist Mathematics

6D WORKED

Example

11

d hca

WORKED

Mat

Example

12i

Partial fractions

Mat

d hca

WORKED

Example

Antidifferentiation

12ii

Antidifferentiation using partial fractions

1 Find the values of a, b and c in the following identities. a ax + b(x − 1) = 3x − 2 b a(x + 2) + b(x − 3) = x − 8 c a(x − 4) + b = 3x − 2 d a(3x + 1) + b(x − 2) = 5x + 4 e a(2 − 3x) + b(x + 5) = 9x + 11 f a(x + 2) + bx = 2x − 10 g a + b(x + 2) + c(x + 2)(x + 3) = x2 + 4x − 2 h a(x + 2)(x − 3) + bx(x − 3) + cx(x + 2) = 3x2 − x + 6 2 Express each of the following rational expressions as partial fractions. 1 12 6x a ---------------------------------b ---------------------------------c ---------------------------------( x + 1)( x + 2) ( x – 2)( x + 2) ( x + 3)( x – 1) 3x x+3 x + 20 d ---------------------------------e ---------------------------------f ---------------------------------( x – 2)( x + 1) ( x + 2)( x + 3) ( x – 4)( x + 4) 5x – 26 4x + 5 x+4 g ------------------2h ------------------2i ------------------x( x – 2) ( x – 5) ( x + 2) 7x – 4 8x – 10 9x – 11 j ---------------------------------k ------------------------------------l ------------------------------------( x – 2)( x + 3) ( 2x + 1 ) ( x – 3 ) ( 3x – 2 ) ( x + 1 ) 11 – 3x 12 – 2x m ---------------------------------n --------------------------------(2 – x)( x + 3) (1 – x)(3 – x) 3 Find the antiderivative of each rational expression in question 2. 4 multiple choice 5x + 10 a b a If ----------------------------2- = ------------ + ----------- , then: x+6 4–x 24 – 2x – x A a = 2, b = 3 B a = −2, b = −3 C a = 3, b = 2 D a = −2, b = 3 E a = 1, b = −1 5x + 10 b Hence ----------------------------2- dx is equal to: 24 – 2x – x A 2loge(x + 6) − 3loge(4 − x) + c B −2loge(x + 6) − 3loge(4 − x) + c C 3loge(x + 6) + 2loge(4 − x) + c D 3loge(x + 6) − 2loge(4 − x) + c log e ( x + 6 ) E --------------------------4–x

∫

5 multiple choice 10 The antiderivative of – ---------------------- is equal to: 2 x +x–6 A 2loge(x + 3) − loge(x − 2) + c

x+1 B 2loge ------------ + c x–6

x+3 C 2loge ------------ + c x–2 E loge(x + 1) − 2loge(x − 6) + c

D loge(x + 3) − 2loge(x − 2) + c

Chap 06 SM Page 239 Thursday, October 12, 2000 10:59 AM

Chapter 6

WORKED

Example

13a

WORKED

Example

13b

Integral calculus

239

6 Antidifferentiate each of the following rational polynomials by first expressing them as partial fractions. 5x – 4 3x + 10 x+3 a ----------------b ---------------------c -------------------------2 2 2 x –x–2 x + 2x x + 3x + 2 5x – 7 6x – 1 x + 16 d ------------------------e -------------------------f -------------------------2 2 2 x – 4x + 3 x + 7x + 6 x – 5x – 6 7x + 1 5x 7x + 9 g -------------h -------------i ----------------------------2 2 2 x –1 2x – 3x – 2 x –9 x+4 4 16 – 2x j ----------------------------k ----------------------------l -------------22 2 2x – 5x + 2 4–x 3x + 7x – 6 x + 13 3x – 4 m ----------------2n --------------------------2 5 + 4x – x 16 – x 7 By first simplifying the rational expression using long division, find the antiderivative of each of the following expressions. x–1 x+3 x2 – 1 a -----------b -----------c ----------------x+5 x–2 x 2 + 3x x 2 + 2x + 4 x2 + x + 4 x2 – x d -------------------------f ------------------------e ---------------------------------2 ( x + 3)( x + 1) x – 4x x 2 – 2x – 3 x 2 + 3x – 2 x 3 + 4x – 13 x 3 + 4x 2 – x ----------------------------g -------------------------i --------------------------------h ( x + 2)( x + 1) x2 – 4 x 2 – 4x – 5 3 2 2 2x + x – 5 x –x+2 2x 2 – 9x + 7 -----------------------------------------------------j ---------------------------k l x2 – 1 x 2 + 2x + 1 x 2 – 6x + 9 8 Evaluate the following integrals in terms of x. 9x + 8 4–x ---------------------------------- dx -------------------- dx a b ( x – 3)( x + 4) x( x + 2) x2 + 3 x 2 + 3x – 4 -------------dx d ---------------------------------- dx e 2 ( x – 4)( x + 2) x –9 x 3 + x 2 – 4x 4x 2 + 6x – 4 ----------------------------- dx ---------------------------- dx g h 2 x 2 – 4x + 4 2x – x – 6 4x – 2 -------------- dx j x2 + 9

∫ ∫ ∫ ∫

Challenge 5x 2 + 2x + 17 --------------------------------------------------- dx k ( x – 1)( x + 2)( x – 3) x 2 + 8x + 9 m -----------------------------------2-dx ( x – 1)( x + 2)

∫ ∫

∫ ∫ ∫

c f i

2

2

2

2

x 2 + 18x + 5 ---------------------------------------------------dx ( x + 1)( x – 2)( x + 3) x 2 + 5x + 1 -----------------------------------dx ( x2 + 1 )( 2 – x )

∫ n ∫ l

5( x + 1)

- dx ∫ ------------------x – 25 x + 4x + 1 - dx ∫ -------------------------x + 6x – 7 x+1 --------------dx ∫x + 4

6 9 a If f ′( x ) = ------------- and f(2) = 3loge 2, find f(x). x2 – 1 b State the domain of f(x). x2 + 1 - and g(4) = 4 − loge 5. 10 a Find g(x) if g′ ( x ) = ------------------------x 2 – 2x – 3 b State the domain of g(x).

Chap 06 SM Page 240 Thursday, October 12, 2000 10:59 AM

240

Specialist Mathematics

Definite integrals The quantity

∫

b

∫ f ( x ) dx is called the ‘indefinite integral of the function f(x)’. However,

f ( x ) dx is called the ‘definite integral of the function f(x)’ and is evaluated using

a

the result that:

∫

b

a

f ( x ) dx = [ F ( x ) ] ab

= F(b) − F(a) where F(x) is an antiderivative of f(x). The definite integral

∫

b

f ( x ) dx can be found only if the integrand, f(x), exists for

a

all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

WORKED Example 14 For each of the following integrals, state: i the domain of the integrand ii whether the integral exists. a

∫

2

–2

1 ------------------ d x 9 – x2

b

∫

4

0

2 ----------------------------------- d x ( x – 1)( x + 3)

THINK a i

WRITE 9 – x2

For the integrand to exist, must be greater than 0.

2

Solve the inequation for x.

x2 < 9 –3 < x < 3

3

State the domain.

The domain is (−3, 3).

ii The integral exists for all values of x between the terminals −2 and 2. b i

a i The integrand exists if

1

1

The integrand does not exist for x = −3 and 1, as these values make the denominator equal to zero.

2

State the domain.

ii The integral does not exist for all values of x between the terminals 0 and 4 (as 1 lies in the interval).

9 – x2 > 0 .

ii The integral exists. b i x ≠ –3, 1

Domain is R\{–3, 1}. ii The integral does not exist.

When using substitution to evaluate definite integrals there is no need to return to an expression in terms of x providing the terminals are expressed in terms of u. In fact it is mathematically incorrect to show the integral in terms of u but with terminals in terms of x. Therefore when using a substitution, u = f(x), the terminals should also be adjusted in terms of u.

Chap 06 SM Page 241 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

241

WORKED Example 15 Use an appropriate substitution to express each of the following definite integrals in terms of u, with the terminals of the integral correctly adjusted. 3 6 x x ------------- dx ---------------- d x a b 2 x–2 2 x –1 3

∫

∫

THINK

WRITE

a

a Let u = x2 − 1.

1

Antidifferentiate the integrand by letting u = x2 − 1 so the derivative method can be applied.

2

du Find ------ . dx

3

Express dx in terms of du.

du or dx = -----2x

Adjust the terminals by finding u when x = 2 and x = 3.

When x = 2, u = 22 − 1 =3 When x = 3, u = 32 − 1 =8

4

du ------ = 2x dx

Therefore the integral is 5

Rewrite the integral.

∫

8

3

6

b

1

Simplify the integrand. Antidifferentiate the integrand by using the linear substitution u = x − 2.

∫

=

x du --- × -----u 2x 8 3

1 ------ du 2u

b Let u = x − 2.

2

du Find ------ . dx

3

Express dx in terms of du.

or dx = du

4

Express x in terms of u.

x=u+2

Adjust the terminals by finding u when x = 3 and x = 6.

When x = 3, u = 3 − 2 =1 When x = 6, u = 6 − 2 =4

5

du ------ = 1 dx

Therefore the integral is 6

7

Rewrite the integral.

Simplify the integrand.

∫

4

1

=

u+2 ----------- du 1 --2 u

∫

4

1

1

1

 u --2- + 2u – --2- du  

Chap 06 SM Page 242 Thursday, October 12, 2000 10:59 AM

242

Specialist Mathematics

WORKED Example 16 Evaluate the following definite integrals. a

∫

2 0

x–2 --------------------------- dx x2 + 5 x + 4

b

∫

π --2

cos x 1 + sin x d x

0

THINK

WRITE

a

a

1

Write the integral.

2

Factorise the denominator of the integrand.

3

Express in partial fraction form with denominators x + 1 and x + 4.

4

5

Express the partial fractions with the original common denominator. Equate the numerators.

6

Let x = −1 to find a.

7

Let x = −4 to find b.

∫

2

x–2 -------------------------- dx 2 x + 5x + 4

0

x–2 x–2 - = ---------------------------------Consider: -------------------------2 ( x + 1)( x + 4) x + 5x + 4 a b = ------------ + -----------x+1 x+4 a( x + 4) + b( x + 1) = ---------------------------------------------x 2 + 5x + 4 x − 2 = a(x + 4) + b(x + 1) Let x = −1, −3 = 3a a = −1 Let x = −4, −6 = −3b b=2 2

8

b

∫ = ∫

So

Rewrite the integral in partial fraction form.

0 2 0

x–2 -------------------------- dx x 2 + 5x + 4 –1 2 ------------ + ------------ dx x+1 x+4

9

Antidifferentiate the integrand.

= [−loge(x + 1) + 2loge(x + 4)]20

10

Evaluate the integral.

= [−loge3 + 2loge6] − [−loge1 + 2loge4] = −loge3 + 2loge6 − 2loge4

11

Simplify using log laws.

= 2loge1.5 − loge3 = loge2.25 − loge3 = loge0.75 (or approx. − 2.88)

1

Write the integral.

b

∫

π --2

cos x 1 + sin x dx

0

2

Let u = 1 + sin x to antidifferentiate.

3

du Find ------ . dx

4

Express dx in terms of du.

Let u = 1 + sin x du ------ = cos x dx du or dx = -----------cos x

Chap 06 SM Page 243 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK 5

243

WRITE When x = 0,

u = 1 + sin 0 =1 π π When x = --- , u = 1 + sin --2 2 =1+1 =2

Change terminals by finding u when

π x = 0 and x = --- . 2

6

Integral calculus

Simplify the integrand.

So

∫

π --2

cos x 1 + sin x dx

0

= =

∫

2

∫

2

1 --- du ( cos x )u 2 -----------cos x 1 1 ---

u 2 du

1

3 2 2 --2--- u 3 1

7

Antidifferentiate the integrand.

=

8

Evaluate the integral.

----2 2 2 2 = --- × 2 – --- × 1 3 3

3

3

4 2 2 = ---------- – --3 3 4 2–2 ------------------3

or

WORKED Example 17

By using the substitution x = sin θ, evaluate

∫

1 --2

1 – x2 d x .

0

THINK

WRITE

1

Let x = sin θ.

2

dx Find ------ . dθ

3

Make dx the subject.

or dx = cos θ dθ

Change the terminals by finding θ when x = 1--2- and x = 0.

= sin θ π θ = --6 When x = 0, 0 = sin θ θ=0

4

Let x = sin θ. dx ------ = cos θ dθ When x = 1--2- ,

1 --2

Continued over page

Chap 06 SM Page 244 Thursday, October 12, 2000 10:59 AM

244

Specialist Mathematics

THINK 5

WRITE

∫

Simplify the integrand.

1 --2

1 – x 2 dx

0

= = =

∫

π --6

0 π --6

2

1 – sin θ cos θ d θ

∫

cos θ cos θ d θ

∫

cos 2 θ d θ

0 π --6 0

6

Replace cos θ by its identity

=

2

1 --- (1 2

+ cos 2θ). =

7

Antidifferentiate the integrand.

8

Evaluate the integral.

π --6 1 --- ( 1 2 0 π --1 6 --(1 2 0

∫

∫

+ cos 2 θ ) d θ + cos 2 θ ) d θ

1 --2

θ + 1--2- sin 2 θ

=

1 --2

 --π- + 1--- sin 6 2

=

1 --2

π 1 3 --- + ---  ------- 6 2 2 

=

π --6 0

π --- – ( 0 + 1--2- sin 0 ) 3

3 π = ------ + ------12 8

Finding the numeric integral at the

Graphics Calculator tip! HOME screen

To find the value of a definite integral, press MATH and select 9:fnInt(. Then type in the integrand, the function variable, the lower terminal and the upper terminal. Press ENTER to evaluate the integral.

Alternatively, if the function is already in Y1, press MATH , select 9:fnInt(, complete 9: fnInt(Y1,X,0, 2) and press ENTER . (Remember that to insert the symbol Y1, press VARS , select Y–VARS and 1:Function, then 1:Y1 (similarly for any Y variable). 2

1 The screen shows both methods for

- dx (Worked example 16a). ∫ -------------------------x + 5x + 4 0

π

2 To estimate

∫ 2 cos 0

2

x–2

2

x --- dx, press MATH , select 9:fnInt( and complete by entering 2

2(cos(X ÷2)) ,X,0,π) and pressing ENTER . 2

Chap 06 SM Page 245 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

245

A handy trick to use, if the answer is a simple fraction, is to press MATH , select ▲

1:

Frac and press ENTER — but it doesn’t work in this case. If the answer could possibly be a fractional multiple of π, first try dividing by π then ▲

pressing MATH , selecting 1: Frac and pressing ENTER . In this case, the answer is just π itself. (Don’t expect this trick to always work!)

remember remember 1.

∫

b

f ( x ) dx = [ F ( x ) ] ab

a

= F(b) − F(a), where F(x) is an antiderivative of f(x).

2. The definite integral

∫

b

f ( x ) dx can be found only if the integrand, f(x), exists

a

for all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

6E WORKED

Example

14

Definite integrals

1 For each of the following definite integrals

∫

b

f ( x ) dx , state i the maximal domain

a

of the integrand f(x) and ii whether the integral exists. a

d g j m

∫

2 1

∫ ∫ ∫

∫

1 -------------- dx 9 – x2

b

dx -------------21+x

e

2 –1

1 –1 3 ------2 0

4x + 10 -------------------------- dx x 2 + 5x + 6

h

dx ----------------4x 2 + 9

k

2

1   x + ----------dx  x – 2 0

n

∫ ∫ ∫ ∫

1 0 4

1 2 0 0 –1

–1 ------------------ dx 4 – x2 2 --- dx x

f i

dx --------------------1 – 9x 2

l

∫ (e

x

+ e – x ) 2 dx

5

∫

3

∫

2

1

1 ------------------2- dx ( x – 1)

3

0

c

∫

3

∫

2

1

1 --2

dx --------------------16 – x 2 dx -------------------x( x + 1) 3x + 2 ----------------------------dx x 2 – 8x + 12 3 ---

( 2x – 1 ) 2 dx

Chap 06 SM Page 246 Thursday, October 12, 2000 10:59 AM

246

Specialist Mathematics

2 Evaluate the integrals in question 1 provided that the integrand, f(x), exists for all values within the domain of the integral.

Mat

d hca

Integrator

3 multiple choice The definite integral

∫

2

2x 2 x 3 + 1 dx can be evaluated after substituting u = x3 + 1.

0

a The integral will then be equal to: A

D

∫

2

∫

2

0

0

2 u ---------- du 3 2x 2

B

u du

E

∫ ∫

9

3 u+1 ------------------- du 2

0

C

∫

9

2 u ---------- du 3

1

3 ---

7

4u 2 -------- du 9

0

b The value of the integral is: B 8---------29

A 11 5--9-

D 12 4--9-

C 9

E 10 10 – 1

4 multiple choice a

π --2

∫

0

cos x ------------------------- dx can be evaluated by first making the substitution: 1 + sin x

A u = sin x D u = cot x

B u = cos x E u = 1 + sin x

C u =

1 + sin x

b The integral will then be equal to:

∫

A

1

1 2

– ---

u du

B

0

c

WORKED

Example

15

∫

0

1 --2

u du

C

1

∫

2

1 2

– ---

u

du

D

0

When evaluated, the integral is equal to: A 2 C −2 B 2 2–2

∫

2

1 2

– ---

u

du

E

∫

1

1 + u du

0

1

E

D 2 2

2 --3

5 By choosing an appropriate substitution for u, express the following integrals in terms of u. (Do not forget to change the terminals.) a

∫

2

∫

4

x 2 ( 2 + x 3 ) dx

b

0

d

( x – 1 ) x 2 – 2x dx

e

2

g

j

∫ ∫

3 1

π --2

m

∫

π --2

1

∫

4x --------------- dx 2 x – 32

2

x x – 1 dx

c

log e x ------------ dx x

h

cos x sin x dx

k

∫ ∫

π --2 π --3 π --4

sin x e cos x dx

cos 3 x dx

n

∫

0

x x 2 + 1 dx

f

i

∫

3 0

x2 ---------------- dx x+1

1

∫ x(1 – x)

10

dx

0

tan 3 x sec 2 x dx

0 1

∫

1

0

1

0

π

∫

2

l

∫

π --2

0

ex ------------------ dx ex + 1

6 Evaluate each of the integrals in question 5.

x sin x 2 dx

Chap 06 SM Page 247 Thursday, October 12, 2000 10:59 AM

Chapter 6

Example

16

7 Evaluate the following definite integrals. a

∫

0

∫

–2

2

b

4xe x dx

c

d

2 x + 3 dx

–3

e

∫

π --3

f

sin x cos 4 x dx

0

g

∫

1

–1

i

∫

1

–1

k

m

∫ ∫

π --2 π --4 1

–1

o

1

h

–1 ---------------------------2- dx 4 – ( x – 1)

j

cot x dx

l

2 1 )e x + x

∫

2

Example

∫

0

2x 2 + 7x dx

–1 2

∫

x+1 ---------------- dx 1–x x+5 -------------------------- dx 2 x + 4x + 3

0

- dx ∫ (--------------------------x – 3) + 1 1

2

0

∫

π

sin 3 x cos 2 x dx

0

2x 2 ------------- dx x2 + 1

∫ ( 2x + 5

∫ ( 4x + 7 )

3

1 ------------------ dx 4 – x2

n

∫ ∫

6 5

π --2

3x – 10 ----------------------------dx x 2 – 7x + 12 2 sin 2x cos x dx

0

p

dx

0

q

1

0

–2

WORKED

247

x2 ---------------- dx x–1

8 By substituting x = sin θ, evaluate

r

∫ ∫

π --3 π --4 4

3

∫

1

( 2 + tan 2 x ) dx 2x 3 + x 2 – 2x – 4 ---------------------------------------- dx x2 – 4

1 – x 2 dx .

0

17

9 By substituting x = 2sin θ, evaluate

∫

3

4 – x 2 dx .

0

10 By making the substitution x = tan θ, evaluate

1

∫

0

11 If

∫

a

∫

a

∫

a

∫

a

0

12 If

0

13 If

4 -------------2- dx = π , find the value of a. 1+x 4 -------------2- dx = – log e 3 , find a. 4–x 3 x + 1 dx = 6 3 , find a.

–1

14 If

dx --------------------. ( 1 + x2 )2

–a

ET SHE

1 π ------------------ dx = --- , find a. 2 2 4–x

Work

WORKED

Integral calculus

6.1

Chap 06 SM Page 248 Thursday, October 12, 2000 10:59 AM

248

Specialist Mathematics

You can check your answers by using the Mathcad file ‘Integrator’ found on the Maths Quest CD-ROM.

Mat

d hca

Integrator

Applications of integration In this section, we shall examine how integration may be used to determine the area under a curve and the area between curves.

Areas under curves You will already be aware that the area between a curve which is above the x-axis and the x-axis itself is as shown in the diagram at right. Area =

∫

b

Further, the area between a curve which is below the x-axis, and the x-axis itself, is as shown in the second diagram.

∫ g( x ) dx = ∫ g( x ) dx

Area = –

y = f (x)

f ( x ) dx

a

b

y

0

a

b

x

y a 0

b x y = g(x)

a

b

a

The modulus is required here since, for a curve segment that lies below the x-axis, the integral associated with that curve segment is a negative number. Area is a positive number and in this case the integral is negative.

Chap 06 SM Page 249 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

Similarly, the area between a curve and the y-axis can be found if the rule for the curve is expressed as a function of y, that is, x = f (y). Area =

∫

b

249

y b

x = f (y)

a

f ( y ) dy (integral measures to the right of the

a

x

0

y-axis are positive) or Area = –

∫

b

y

g ( y ) dy (integral measures to the left of the

b

a

x = g (y)

y-axis are negative) Area =

∫

b

a

g ( y ) dy

If the graph crosses the x-axis, then the areas of the regions above and below the x-axis have to be calculated separately. In this case the x-intercepts must be determined. In the figure at right a single intercept, c, is shown. b

c

y y = f (x)

c

c

a

Similarly the shaded region in the figure at right has an area given by:

y b c

c

∫ f ( y ) dy + ∫ f ( y ) dy = ∫ f ( y ) dy – ∫ f ( y ) dy

Area =

c

x

b

a

b

b

0 c

a

c

∫ f ( x ) dx + ∫ f ( x ) dx = ∫ f ( x ) dx – ∫ f ( x ) dx

Area =

x

0

a

x = f (y)

a

b

c

c

a

0 a

WORKED Example 18 2 log e x If y = ----------------- , find: x a the x-intercepts

b the area bounded by the curve, the x-axis and the line x = 3.

THINK a

1 2

b

1

For x-intercepts, y = 0, when 2loge x = 0. Solve for x. Sketch a graph showing the region required. (A graphics calculator may be used.)

WRITE a x-intercepts occur when 2loge x = 0. That is, x = 1. b y 2 loge x y = ———— x

0 1

3

x

Continued over page

x

Chap 06 SM Page 250 Thursday, October 12, 2000 10:59 AM

250

Specialist Mathematics

THINK 2 3

WRITE

Express the area as a definite integral. Antidifferentiate by letting u = loge x to apply the derivative method.

4

du Find ------ . dx

5

Make dx the subject.

6

Express the terminals in terms of u.

∫

3

2 log e x ---------------- dx x 1 Let u = loge x.

Area =

du 1 ------ = --x dx or dx = x du When x = 1, u = loge1 =0 When x = 3, u = loge3 Area =

∫

log e 3

∫

log e 3

0

7

=

Simplify the integrand.

2u ------ x du x 2u du

0 log e 3

8

Antidifferentiate the integrand.

= [ u 2 ]0

9

Evaluate the integral.

= [ ( log e 3 ) 2 ] – [ 0 2 ] = (loge3)2

10

State the area.

The area is (loge3)2 or approximately 1.207 square units.

A graphics calculator should be used here to verify the result.

Finding the numeric integral at the

Graphics Calculator tip! GRAPH screen

To find the area under a curve between two x-values, first graph the curve by entering its equation as Y1 in the Y= menu. 2log e x Consider y = ------------------ in worked example 18. Press Y= and type in (2ln(X))÷X at Y1. x Then press GRAPH . To find the area bounded by this curve and the x-axis between x = 1 and x = 3, press 2nd [CALC] and select 7: ∫f(x) dx. Type in 1 for the lower value (press ENTER ) and 3 for the upper value (press ENTER ). Compare this result to that obtained in worked example 18.

Chap 06 SM Page 251 Thursday, October 12, 2000 10:59 AM

Chapter 6

251

Integral calculus

WORKED Example 19

Examine the figure at right. a Express the rule as a function of y. b Find the area of the shaded section. THINK

WRITE

a

a

Write down the rule.

2

Square both sides of the equation. Add 1 to both sides to make x the subject.

1

Express the area between the curve and the y-axis in integral notation.

2

Antidifferentiate by rule.

3

Evaluate the integral.

y = x–1

2

1

3

b

y

y= x–1 y2 = x − 1 or x = y2 + 1

b Area =

2

∫ (y

2

x

0 1

+ 1 ) dy

0

[ 1--3- y 3 + y ]02 = [ 8--3- + 2 ] – [ 0 + 0 ] =

= 4 2--34

The area is 4 2--3- square units.

State the area.

The shaded area shown in the figure in worked example 19 could also have been calculated relative to the x-axis by subtracting the area between the curve and the x-axis from the area of the rectangle as shown in the figure at right. That is: Area = ( 5 × 2 ) –

∫

y y = x–1

2

x 5

0 1

5

x – 1 dx

1

Using symmetry properties In some problems involving area calculations, use of symmetry properties can simplify the procedure.

WORKED Example 20

Find the area inside the ellipse in the figure at right. THINK WRITE y2 x 2 + ----- = 1 1 Write the equation. (The ellipse is 4 symmetrical about the x-axis and y-axis and so finding the shaded area in the figure allows for the total enclosed area to be determined.) y2 ----- = 1 − x2 2 Express the relation as a function of x 4 for the top half of the ellipse. y2 = 4(1 − x2)

y 2

2

x2 + —y4 = 1 0 1

–1

x

–2

y = 2 1 – x 2 is the rule for the top half of the ellipse. (y = – 2 1 – x 2 is the bottom half.) Continued over page

Chap 06 SM Page 252 Thursday, October 12, 2000 10:59 AM

252

Specialist Mathematics

. THINK 3

4

WRITE

Write the integral that gives the area in the first quadrant (a quarter of the total area). Express the total area as four times this integral.

Area in the first quadrant =

∫

1

2 1 – x 2 dx

0

∫ =8 ∫

Total area of ellipse = 4

1

2 1 – x 2 dx

0 1

1 – x 2 dx

0

8

To antidifferentiate, let x = sin θ. dx Find ------ . dθ Make dx the subject. Express the terminals in terms of θ.

9

Rewrite the integral in terms of θ.

5 6 7

10

Simplify the integrand using identities.

Let x = sin θ.

dx ------ = cos θ dθ or dx = cos θ dθ When x = 0, sin θ = 0 θ=0 When x = 1, sin θ = 1 π θ = --2 Area = 8 =8

∫

∫

π --2

0 π --2

1 – sin 2 θ cos θ dθ cos 2 θ d θ

0

=8 =4

π --2 1 --- ( 1 2 0

∫ ∫

π --2

+ cos 2 θ ) d θ

( 1 + cos 2 θ ) d θ

0

π ---

11

Antidifferentiate the integrand.

= 4 [ θ + 1--2- sin 2 θ ] 02

12

Evaluate the integral.

π 1 = 4 --- + --- sin π – [ 0 + 1--2- sin 0 ] 2 2

13

State the area.

π = 4  --- + 0 2  = 2π The exact area is 2π square units.

Areas between curves When finding the areas between two curves that intersect, it is necessary to determine where the point of intersection occurs. In the figure at right, two functions, f and g, intersect at the point P with x-ordinate c. The area contained within the envelope of the two functions bounded by x = a and x = b is given by:

y

a

P

f (x)

0 c

g (x) x

b

Chap 06 SM Page 253 Thursday, October 12, 2000 10:59 AM

Chapter 6

Area =

∫

c

[ g ( x ) – f ( x ) ] dx +

a

Integral calculus

b

∫ [ f ( x ) – g( x ) ] dx c

Similarly, areas between curves can also be found relay tive to the y-axis. x = g (y)b Area =

253

x = f (y)

b

∫ [ g( y ) – f ( y ) ] dy

a

a

Note that on the interval [a, b], g(y) ≥ f (y) and hence the integrand is g(y) − f (y) and not f (y) − g(y).

0

When an area between a curve and the x-axis (or between curves) gives an integrand which cannot be antidifferentiated, it may be possible to express the area relative to the y-axis, creating an integrand which can be antidifferentiated.

WORKED Example 21

Find the area bounded by the curves y = x2 − 2 and y = 2x + 1. THINK 1

2

3

WRITE

Check on a graphics calculator to see if the curves intersect. If they do, solve x2 − 2 = 2x + 1 to find the x-ordinate of the point or points of intersection for the two curves.

x2 − 2 = 2x + 1 x − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = 3 and x = −1 The curves intersect at x = 3 and x = −1.

Express the area as an integral. (Use , as without a graph we cannot always be sure which function is above the other. Here is a valuable use for the graphics calculator.)

Area =

Simplify the integral.

2

∫

3

∫

3

[ x 2 – 2 – ( 2x + 1 ) ] dx

–1

=

( x 2 – 2x – 3 ) dx

–1

4

Antidifferentiate by rule.

=

[ 1--3- x 3 – x 2 – 3x ]–31

5

Evaluate the integral.

=

[ ( 9 – 9 – 9 ) – ( – 1--3- – 1 + 3 ) ]

= – 9 – 1 2--3= – 10 2--3= 10 2--36

State the solution.

The area bounded by the two curves is 10 2--3- square units.

x

Chap 06 SM Page 254 Thursday, October 12, 2000 10:59 AM

254

Specialist Mathematics

Showing and finding the area

Graphics Calculator tip! bounded by two curves Consider using the TI calculator for worked example 21. 1. To graph the two curves with equations y = x2 – 2 and y = 2x + 1 enter Y1= X2 – 2 and Y2= 2X + 1. Then press GRAPH . Use TRACE to locate the points of intersection. Adjust the WINDOW settings if necessary. 2. To show the area bounded by the two curves, press Y= , position the cursor to the left of the Y1 symbol and press ENTER successively to obtain the ‘shade below’ style. Repeat for Y2 to obtain the ‘shade above’ style. Press GRAPH . The required area is shown unshaded.

3. To determine the value of the area bounded by the curves on the required interval (in this case, between x = –1 and x = 3), press MATH , select 9 and complete 9: fnInt(Y2–Y1,X,–1,3) and press ENTER . Remember, to insert Y1, press VARS and select Y–VARS, 1:Function and 1:Y1 (or 2:Y2 to enter Y2). Note that in this case we are subtracting Y1 from Y2 (seen by viewing the graph). However, if it is entered the opposite way, it only produces the negative of the required answer.

remember remember

1. The area between a curve f (x), the x-axis and lines x = a and x = b is given by: Area =

∫

b

∫

b

f ( x ) dx = F ( b ) – F ( a ) where F(x) is the antiderivative of f (x).

a

2. Area measures can also be evaluated by integration along the y-axis. The area between a curve f (y), the y-axis and lines y = a and y = b is given by: Area =

f ( y ) dy = F ( b ) – F ( a ) where F(y) is the antiderivative of f (y).

a

3. If an area measure is to be evaluated over the interval [a, b] and the curve crosses the x-axis at x = c between a and b, then the integral has to be decomposed into two portions. Area =

∫

c

a

f ( x ) dx +

∫

b

f ( x ) dx = F ( c ) – F ( a ) + F ( b ) – F ( c )

c

4. The area bounded by two curves f (x) and g(x) where f (x) ≥ g(x) and the lines x = a and x = b is given by: Area =

b

∫ [ f ( x ) – g( x ) ] dx = F ( b ) – G( b ) – F ( a ) + G( a ) a

5. Where possible use a graphics calculator to draw the function or functions to determine whether the integrals have to be decomposed into portions and to check and verify the correct use of the modulus function.

Chap 06 SM Page 255 Thursday, October 12, 2000 10:59 AM

Chapter 6

6F

255

Integral calculus

Applications of integration

For the following problems, give exact answers wherever possible; otherwise give answers to an appropriate number of decimal places. (Use a graphics calculator to assist with, or verify, any graphing required.)

WORKED

18

1 For each of the following curves find: i the x-intercepts ii the area between the curve, the x-axis and the given lines. x , x = 0 and x = 9

1 b y = x − ----2- , x = 1 and x = 2 x

y = x x – 1 , x = 2 and x = 5

3x – 2 - , x = 3 and x = 4 d y = -------------x2 – 4

a y= c

1 e y = ------------------ , x = 1 and x = 4 – x2

f

3

π g y = 2x cos x2, x = – --- and x = 0 3 WORKED

Example

19

Math

cad

Example

Definite integral – graph

π y = cos2x, x = 0 and x = --2

ex h y = --------------x , x = 0 and x = 1 2+e

2 For each of the graphs below: i express the relationship as a function of y (that is, make x the subject of the rule) ii find the magnitude of the shaded area between the curve and the y-axis. a b c y y 2 y y = (x – 1)

2

–π 2 –π 3

4

y= x

–1

1 x

0

d

e

y y = logex

2

–π 2

–1

x

0 1

g

f

y

π

y

y = x3

8

y = Cos–1 x 0 1

x 0

y

y = Tan–1 x

–π 2 –π 4

0

x

x

0 1

– –π 2

x

–1 0

y = Sin–1 x

x

Chap 06 SM Page 256 Thursday, October 12, 2000 10:59 AM

256 WORKED

Example

20

Specialist Mathematics

3 Find the magnitude of the shaded areas on each graph below. a b y c y y = x2

0 0

d

0

f

y

y = loge x

x2 –– 9

1

x

+

y2

x

1

y

=1

1

3 x

0

–3

1 ––––– 4 + x2

x

2

e

e2

y=

–1

y

0 1

–1 4

x

2

y

y2 = x

0

–1

–π 2

x

π

y = sin3x

4 multiple choice a The definite integral that correctly gives the area bounded by the curve y = 4x − x2 and the x-axis is: 2

A

1

∫ ∫ ( 4x – x ) dx ( 4x – x 2 ) dx

B

0

4

D

∫ ∫ ( 2x

0

( 4x – x 2 ) dx

C

0

0

2

E

0

∫ ( 4x – x ) dx 2

4

2

2

– 1--3- x 3 ) dx

b The area, in square units, is equal to: A 10 2--3-

B 2 1--3-

C 5 1--3-

E −5 1--3-

D 8

5 multiple choice a Which of the graphs below correctly shows the area bounded by the curve y2 = x + 1 and the y-axis? A B C y y y y2 = x + 1

y2 = x + 1

2

y2 = x + 1 1

D

–1

x

0

E

y

–2

y y2 = x + 1

1

y2 = x + 1 0

x

0 x

0

–1

x

0

x

–1

b The definite integral which gives the area bounded by y2 = x + 1 and the y-axis is: 1

A

∫ ∫ (y

( y 2 – 1 ) dy

B

0

1

D

0

∫ ∫

0

( y 2 – 1 ) dy

–1

2

+ 1 ) dy

E

∫ (y 1

1

0

0

C 2

x – 1 dx

2

– 1 ) dy

Chap 06 SM Page 257 Thursday, October 12, 2000 10:59 AM

Chapter 6

c

The value of the area, in square units, is equal to A

WORKED

Example

257

Integral calculus

2 --3

B 2 2--3-

C 1 1--3-

D 5 1--3-

E 2

6 Find the area bounded by the graph with equation y = (x − 2)2(x + 1) and the x-axis.

21

7 Find the area bounded by the graph with equation y2 = x + 4 and the y-axis. 8 a Show that the graphs of f (x) = x2 − 4 and g(x) = 4 − x2 intersect at x = −2 and x = 2. b Find the area bounded by the graphs of f (x) and g(x).

10 a On the same axis sketch the graphs of y = 9 – x and y = x + 3. b Find the value of x where the graphs intersect. c Hence find the area between the curves from x = −1 to x = 2. 11 Find the area bounded by the curves y = x2 and y = 3x + 4. 12 Find the area enclosed by the curves y = x2 and y =

x.

13 Find the area bounded by y = e x and y = e−x and the line y = e. 14 Examine the figure at right. a Find the area enclosed by f (x), g(x) and the y-axis. b Find the shaded area. x2

1– e

y 2 1

y2

15 Find the area of the ellipse with equation ----2- + ----2- = 1 . a b Hints:1. Use symmetry properties. 2. Antidifferentiate

g (x) = e x – 1 2 f (x) = ––––– 1 + x2

0

1

2

x

a 2 – x 2 by using the substitution x = asin θ.

x2 y2 16 Find the area between the circle x2 + y2 = 9 and ellipse ----- + ----- = 1 . 9 4 Hint: Make use of symmetry properties. 17 a Sketch the curve y = e x + 2. b Find the equation of the tangent at x = −2. c Find the area between the curve, the tangent and the y-axis. 1–x 18 a Sketch the graph of y = ------------ . x+1 b Find the area bounded by this curve and the x- and y-axes. 1 19 a Show algebraically that the line y = x does not meet the curve y = ------------------ . 1 – x2 1 b Find the area enclosed by the curve, the lines y = x and x = ------- , and the y-axis. 2

cad

9 a On the same axis sketch the graphs of f (x) = sin x and g(x) = cos x over [0, π]. π b Show algebraically that the graphs intersect at x = --- . 4 c Find the area bounded by the curves and the y-axis.

Math

Area between curves

Chap 06 SM Page 258 Thursday, October 12, 2000 10:59 AM

258

Specialist Mathematics

Volumes of solids of revolution If part of a curve is rotated about the x-axis, or y-axis, a figure called a solid of revolution is formed. For example, a solid of revolution is obtained if the shaded region in figure 1 is rotated about the x-axis. y

y

y y = f(x)

y = f(x)

y = f(x) y

0 a

x

b

0 a

Figure 1

b

x

0 a

x

b

δx Figure 3

Figure 2

The solid generated (figure 2) is symmetrical about the x-axis and any vertical crosssection is circular, with a radius equal to the value of y at that point. For example, the radius at x = a is f (a). Any thin vertical slice may be considered to be cylindrical, with radius y and height δx (figure 3). The volume of the solid of revolution generated between x = a and x = b is found by allowing the height of each cylinder, δx, to be as small as possible and adding the volumes of all of the cylinders formed between x = a and x = b. That is, the volume of a typical strip is equal to π y2 δx. Therefore the volume of the solid contained from x = a to x = b is the sum of all the infinitesimal volumes: x=b

V = lim

δx → 0

=

∑ π y2 δx

x=a

b

∫ πy

2

dx

a

The value of y must be expressed in terms of x so that the integral can be evaluated. From the figure above y = f (x) and thus the volume of revolution of a curve f (x) from x = a to x = b is V = π

b

∫ [ f ( x)]

2

dx .

a

Similarly if a curve is rotated about the y-axis, the solid of revolution shown in the figure at right is produced. The volume of the solid of revolution is likewise V = π

y b

b

∫ [ f ( y)]

2

dy

a

a

For regions between two curves that are rotated about the x-axis: V = π

x = f (y)

b

∫ [ f ( x)]

2

x

0

y

y = f (x) y = g (x)

– [ g ( x ) ] 2 dx

a

0

a

b

x

Chap 06 SM Page 259 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

259

WORKED Example 22

a Sketch the graph of y = 2x and show the region bounded by the graph, the x-axis and the line x = 2. b Find the volume of the solid of revolution when the region is rotated about the x-axis. THINK WRITE a a 1 Sketch the graph. y x = 2 y = 2x 2 Shade the region required.

0

b

1

2

State the integral that gives the volume. (The volume generated is bounded by x = 0 and x = 2.)

bV= π

2

∫ ( 2x )

2

2

∫ 4x

dx

2

dx

0 4 3 2 --- x 3 0

3

Antidifferentiate by rule.

= π[

4

Evaluate the integral.

------ – 0 ] = π [ 32 3

5

x

0

=π

Simplify the integrand.

2

]

32 π = --------3 32 π The exact volume generated is --------- cubic units. 3

State the volume.

Finding the volume of a solid

Graphics Calculator tip! of revolution

Consider the volume of the solid of revolution formed in Worked example 22. 1. The line with equation y = 2x is rotated about the x-axis to form a cone. To graph the line, enter Y1 = 2X and press GRAPH . Press TRACE to locate particular coordinates. 2. To determine the volume of the cone, press MATH , then select 9: fnInt( and insert π Y12, X, 0, 2) and press ENTER . To insert Y1, press VARS and select Y–VARS, 1:Function and 1:Y1 (similarly any Y variable). (Note that π Y12 provides the integrand, X is the variable, 0 and 2 are the terminals of the integral.) 1 3 Can you verify the formula V = --- π r for this cone? What is the radius for this cone? 3 You can try to convert your volume answer to a fraction of π . Press ÷ and 2nd [ π ], ▲

then MATH . Select 1:

Frac and press ENTER . This is also shown in the screen above.

Chap 06 SM Page 260 Thursday, October 12, 2000 10:59 AM

260

Specialist Mathematics

WORKED Example 23

a Sketch the region bounded by the curve y = loge x, the x-axis, the y-axis and the line y = 2. b Calculate the volume of the solid generated if the region is rotated about the y-axis. THINK WRITE a 1 Sketch the graph. (Use a graphics a y calculator if necessary.) y = logex 2 2 Shade the region required. 0

b

1 2 3

4

5

Write the rule y = loge x. Take the exponent of both sides to get y as a function of x. State the function.

b y = loge x ey = elog x ey = x or x = ey

Express the volume in integral notation between y = 0 and y = 2.

2

∫ ( e ) dy = π e dy ∫

So V = π

0 2

Simplify the integrand.

6

Antidifferentiate by rule.

= π[

7

Evaluate the integral.

= π[

8

x

1

y 2

2y

0 1 2y --- e 2 1 4 --- e – 2

]02

1 0 --2

e

]

π = --- ( e 4 – 1 ) 2 π The volume is exactly --- ( e 4 – 1 ) cubic units 2 (or approximately 84.19 cubic units).

State the volume.

remember remember 1. To find the volume of revolution about the x-axis for the function f (x) from x = a to x = b, evaluate the integral: V= π

b

∫ [ f ( x)]

2

dx

a

2. To find the volume of revolution about the y-axis for the function f (y) from y = a to y = b, evaluate the integral: V= π

b

∫ [ f ( y)]

2

dy

a

3. To find the volume of revolution about the x-axis for the region between f (x) and g(x) where f (x) ≥ g(x) from x = a to x = b, evaluate the integral: V= π

b

∫ [ f ( x)] a

2

– [ g ( x ) ] 2 dx

Chap 06 SM Page 261 Thursday, October 12, 2000 10:59 AM

Chapter 6

6G

Integral calculus

261

Volumes of solids of revolution

Give exact answers where possible; otherwise use an appropriate number of decimal places when giving approximate answers. (Use a graphics calculator to check any graphing.) WORKED

22

1 a Sketch the graph of the region bounded by the x-axis, the curve y = 3x and the line x = 2. Solid of b Calculate the volume generated by rotating this region about the x-axis. revolution x c Verify this result by using the standard volume formula for the solid generated.

Math

cad

Example

2 The region bounded by the graph of y = 16 – x 2 and the x-axis is rotated about the x-axis. a Calculate the volume of the solid of revolution generated. b Verify this answer using the standard volume formula. WORKED

23

3 a Sketch the region bounded by the curve y = x – 1 , the y-axis and the lines y = 0 and y = 2. Solid of b Calculate the volume generated when this region is rotated about the y-axis. revolution y

4 Find the volume generated when the area bounded by y = x2 − 1 and the x-axis is rotated about: a the x-axis b the y-axis. 5 For the regions bounded by the x-axis, the following curves, and the given lines: i sketch a graph shading the region ii find the volume generated when the region is rotated about the x-axis. a y = x + 1; x = 0 and x = 2 b y = x ; x = 1 and x = 4 c

y = x2; x = 0 and x = 2

e x2 + y2 = 4; x = −1 and x = 1

π π g y = cos x; x = – --- and x = --2 2

d y2 = 2x + 1; x = 0 and x = 3 2 f y = --- ; x = 1 and x = 3 x

h y = ex + 1; x = −2 and x = −1

6 For each region defined in question 5 (a to f only) find the volume generated by rotating it about the y-axis. 7 multiple choice a The region bounded by the curves y = x2 + 2 and y = 4 − x2 is represented by the graph: A B 2 y y y=x +2

y = x2 + 2

–2

0

2

x

y = 4 – x2

(–1, 3)

(1, 3) x

0 y = 4 – x2

Math

cad

Example

Chap 06 SM Page 262 Thursday, October 12, 2000 10:59 AM

262

Specialist Mathematics

C

D

y y = x2 + 2

(–2, 2)

y

y = x2 + 2

(2, 2) 0 x

0

y = 4 – x2

y = 4 – x2

E y = 4 – x2

x

4

y

y = x2 + 2 (1, 3) x

0 (1, –3)

b The volume generated when the region is rotated about the x-axis is equal to: A π

∫

2

∫

1

∫

1

( 2 – 2x 2 ) 2 dx

B π

0

C π

∫

1

( 2 – 2y ) 2 dy

–1

( 2 – 2x 2 ) 2 dx

D π

( 2 – 2x 2 ) 2 dx

–1

0

E π

∫

1

( 6 – 2x 2 ) 2 dx

–1

c

The volume generated when the region is rotated about the y-axis is equal to: A π

∫

4

∫

4

( 4 – y ) dy + π

3

C π E π

( y – 2 ) dy

B π

2

( 2 – 2y ) dy

2 4

∫

3

∫

4

( y – 2 ) dy + π

3

D π

3

∫ ( 4 – y ) dy 2

4

∫ ( 2y – 2 ) dy 2

∫ ( 2 – 2x ) dx 2

2

8 Find the volume generated when the region bounded by the curves y = x2 and y = −x is rotated about: a the x-axis b the y-axis. 9 Find the volume generated when the area bounded by the curve y = sec x, the line π x = --- and the x- and y-axes is rotated about the x-axis. 4 10 Find the volume generated by rotating the area bounded by y = e2x, the y-axis and the line y = 2 about the x-axis. 11 The area bounded by the curve y = Tan−1x, the x-axis and the line x = 1 is rotated about the y-axis. Find the volume of the solid generated. x2 12 A model for a container is formed by rotating the area under the curve of y = 2 – ----6 between x = −1 and x = 1 about the x-axis. Find the volume of the container.

Chap 06 SM Page 263 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

13 For the graph shown at right: a find the coordinate of A b find the volume generated when the shaded region is rotated about the x-axis c find the volume generated when the shaded region is rotated about the y-axis.

263

y

0

A

y=

1

2

1 ––––– 4 – x2

x

x2 y2 14 What is the volume generated by rotating the ellipse with equation ----- + ----- = 1 about: 4 9 a the x-axis? b the y-axis? 15 Find the volume generated when the region bounded by y = x2 and y = rotated about: a the x-axis b the y-axis.

8x is

16 Find the volume generated by the rotation of the area bounded by the curves y = x3 and y = x2 about: a the x-axis b the y-axis. 17 A hemispherical bowl of radius 10 cm contains water to a depth of 5 cm. What is the volume of water in the bowl? 18 A solid sphere of radius 6 cm has a cylindrical hole of radius 1 cm bored through its centre. What is the volume of the remainder of the sphere? 19 Find the volume of a truncated cone of height 10 cm, a base radius of 5 cm and a top radius of 2 cm. 20 a Find the equation of the circle sketched below. b Find the volume of a torus (doughnut-shaped figure) generated by rotating this circle about the x-axis (give your answer in cm3). y

6 0

4

x

Chap 06 SM Page 264 Thursday, October 12, 2000 10:59 AM

264

Specialist Mathematics

Approximate evaluation of definite integrals and areas When calculating definite integrals or areas that involve integrands which cannot be antidifferentiated using techniques discussed in this chapter, approximation methods can be used. We shall now look at two useful and simple approximation methods: the midpoint rule and the trapezoidal rule.

The midpoint rule The definite integral

∫

b

f ( x ) dx determines the shaded area

y

y = f (x)

a

under the curve below. It can be approximated by constructing a rectangle with height equal to the value of y halfway between x = a and x = b. a+b Area of rectangle = ( b – a ) f  ------------  2  The estimate for the shaded area is improved by increasing the number of intervals, that is the number of rectangles between x = a and x = b. In the figure below, the region from x = a and x = b is broken up into n rectangles. The base width of each rectangle is δx and the height of each individual rectangle is obtained from the midpoint rule. The area of each rectangle is given by the product of the height and the common width δx.

0

a

y

0

y = f (x)

a

y

∫

b–a where: δ x = ------------ , the width of each rectangle n n = the number of intervals and hence rectangles used x0 = a xn = b

WORKED Example 24 4

Estimate

∫ (x

2

+ 2 x ) d x using the midpoint rule and 4 intervals.

0

THINK 1 State f (x). 2 Calculate δx.

WRITE f (x) = x2 + 2x 4–0 δx = -----------4 =1

x

b

0 x0 x1 x2 … xn x

x0 + x1 x1 + x2 xn – 1 + xn f ( x ) dx ≈ δ x f  ---------------- + f  ---------------- + . . . + f  -----------------------  2   2    2 a b

a+b ––– 2

y = f (n)

(a)

So

x

b

(b)

Chap 06 SM Page 265 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

265

Integral calculus

WRITE

3

Find x0, x1, x2, x3, x4.

4

Substitute these values into the midpoint rule.

x0 = 0 x1 = 1 x2 = 2 x3 = 3 x4 = 4 4

5

Evaluate the approximation.

6

State the solution.

So

∫ (x

2

+ 2x ) dx

0

≈ 1[ f (0.5) + f (1.5) + f (2.5) + f (3.5)] = 1(1.25 + 5.25 + 11.25 + 19.25) = 37

The approximate value of the definite integral is 37.

The trapezoidal rule

y

The area under a curve can also be approximated using a trapezium. b–a The area of the trapezium = ------------ × [ f ( a ) + f ( b ) ] . 2

0

By increasing the number of intervals between x = a and x = b, that is, the number of trapezia, the estimate becomes more accurate:

So

∫

b

a

f ( x ) dx

y = f (x)

a

x

b

y y = f (x)

0 x0 x1 x2 x3 … (a)

xn x (b)

b–a ≈ ------------ × { [ f ( a ) + f ( x 1 ) ] + [ f ( x 1 ) + f ( x 2 ) ] + [ f ( x 2 ) + f ( x 3 ) ] + . . . + [ f ( x n – 1 ) + f ( b ) ] } 2 Notice here that the terms f (a) and f (b) occur only once and all other terms such as f (x1) and f (x2) occur twice. Thus an approximation to the area is:

δx Approximate area = ------ [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + . . . + 2 f ( x n – 1 ) + f ( x n ) ] 2 where: n = the number of intervals used b–a δ x = -----------n a = x0 b = xn

Chap 06 SM Page 266 Thursday, October 12, 2000 10:59 AM

266

Specialist Mathematics

WORKED Example 25 4

Estimate

∫ (x

2

+ 2 x ) d x using the trapezoidal rule and four equal intervals.

0

THINK

WRITE

1

State f (x).

2

Calculate δx.

3

Find x0, x1, x2, x3, x4.

4

Substitute these values into the trapezoidal rule.

f (x) = x2 + 2x 4–0 δx = -----------4 =1 x0 = 0 x1 = 1 x2 = 2 x3 = 3 x4 = 4 4

So

∫ (x

2

+ 2x ) dx

0

≈ 1--2- [f (0) + 2f (1) + 2f (2) + 2f (3) + f (4)] 5

Evaluate the approximation.

= 1--2- [0 + 2(3) + 2(8) + 2(15) + 24] = 1--2- (76)

6

State the solution.

= 38 The value of the definite integral is approximately 38.

Compare this answer with that in worked example 24. Which is closest to the exact answer?

WORKED Example 26

Estimate the area under the graph of y = x log e x from x = 1 to x = 5 using two equal intervals and: a the midpoint rule b the trapezoidal rule. THINK WRITE a 1 State f (x). a f (x) = x loge x 5–1 δx = -----------2 Calculate δx. 2 =2 x0 = 1 3 Find x0, x1, x2. x1 = 3 x2 = 5 5 So the area = x log e x dx 4 Substitute the values into the 1 midpoint rule. ≈ 2[ f(2) + f(4)] = 2[2loge2 + 4loge4] 5 Evaluate the estimate of the area. = 4loge2 + 8loge4 The approximate area is 13.863 square units. 6 State the approximate area.

∫

Chap 06 SM Page 267 Thursday, October 12, 2000 10:59 AM

Chapter 6

THINK

WRITE

b

b f (x) = xloge x

1

State f (x).

2

Calculate δx.

3

Find x0, x1, x2.

4

Substitute these values into the trapezoidal rule.

5

Evaluate the estimate for the area.

6

State the approximate area.

5–1 δx = -----------2 =2 x0 = 1 x1 = 3 x2 = 5 So the area =

267

Integral calculus

5

∫ xlog x dx e

1

≈ 2--2- [f (1) + 2f (3) + f (5)] = 1[loge1 + 2(3loge3) + 5 loge5] = 6 loge3 + 5 loge5 The approximate area is 14.639 square units.

remember remember 1. Some functions cannot be integrated using the y y = f (n) techniques covered in this chapter. Two approximation methods are discussed. The midpoint rule involves subdividing the required area into a finite number of rectangles. The trapezium rule involves 0 x0 x1 x2 … xn x subdividing the area into a finite number of trapezia. (a) (b) 2. The midpoint rule: b x0 + x1 x1 + x2 xn – 1 + xn f ( x ) dx ≈ δ x f  ---------------- + f  ---------------- + . . . + f  -----------------------       2 2 2 a

∫

b–a where: δx = ------------ , the width of each rectangle n n = the number of intervals used x0 = a xn = b 3. The trapezoidal rule:

y y = f (x)

0 x0 x1 x2 x3 … (a)

δx f ( x ) dx ≈ ------ [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + . . . + 2 f ( x n – 1 ) + f ( x n ) ] 2 a where: n = the number of intervals used b–a δx = -----------n a = x0 b = xn

∫

b

xn x (b)

Chap 06 SM Page 268 Thursday, October 12, 2000 10:59 AM

268

Specialist Mathematics

Approximate evaluation of definite integrals and areas

6H WORKED

Example

24

1 Find approximations to the following definite integrals using the midpoint rule with four equal intervals. a

∫ ∫

5

dx ----------x–2

3

c

b

Example

sin x dx

0

–1

d

log e x 2 dx

4

Tan –1 x dx

0

–3

WORKED

∫ ∫

π

2 Repeat question 1 using the trapezoidal rule.

25

GC p

am rogr

WORKED

Midpoint Example rule 26a

3 Use the midpoint rule with two equal intervals to estimate the following definite integrals. a

d hca

2

∫ ∫ (x

( x + 2 ) dx

b

Mat

0

Middle boxes

c

Example

∫ ( x – 3 ) dx ∫ ( 16 – x ) dx 2

1

3

– x 2 + 2x ) dx

d

1

WORKED

am rogr

2

4

4

2

0

4 Repeat question 3 using the trapezoidal rule.

GC p

26b Trapezoidal rule

5 multiple choice a Using the midpoint rule and two equal intervals, an estimate for

Mat

d hca

Trapezoidal rule

A 1.4

B 0.8

C 0.9412

D 0.7906

∫

1

1 -------------2- dx is: + 1 x 0 E 0.863

b Compared to the exact answer, the percentage error in answer a is closest to: A 0.7 B 1.9 C 2.5 D 20 E 6.4 6 multiple choice a Using the trapezoidal rule and four equal intervals, an estimate for

π --2 π – --2

∫

is: A 1+ 2

π B --2

π π(1 + 2) π(1 + 2) C ------------------------- D ------------------------- E --4 4 8

b The percentage error relative to the exact answer is closest to: A 61 B 21 C 11 D 5 7 Find the value of

5

∫e 1

a 4 equal intervals b 8 equal intervals.

x

dx using the midpoint rule and:

E 1

cos x dx

Chap 06 SM Page 269 Thursday, October 12, 2000 10:59 AM

Chapter 6

8 Find

∫

4

Integral calculus

269

1 + x 2 dx using the trapezoidal rule and:

0

a 4 equal intervals b 8 equal intervals. 9 Estimate the area under the curve y = loge x from x = 1 to x = 4 using the midpoint rule and: a 3 equal intervals b 6 equal intervals. 10 Estimate the area under the graph of y = Cos−1x from x = −1 to x = 1 using the midpoint rule and 4 equal intervals. 11 Calculate an estimate for the area under the graph of y = 2x between x = 0 and x = 2 using the trapezoidal rule and: a 2 equal intervals b 4 equal intervals. 12 Using the trapezoidal rule with: a 2 equal intervals b 4 equal intervals find the approximate area under the graph of y =

sin x between x = 0 and x = π.

Is there a connection between area and volume of revolution? The area under a curve can be found directly from integration. Likewise, the volume of revolution can also be found from integration. In both cases it is assumed that the curves have functions which can be readily integrated. This investigation examines the relative size of the area produced by a curve and the x-axis, and the volume of revolution produced. Curves will be restricted to those of the type f (x) = ax n. 1 Consider the function f (x) = ax n, where a = 1--2- , 1, 2 and 4 and n = 1. Find the area enclosed by the curve, the x-axis and the line x = 1 for each of the four values for a. In what manner is the area dependent on the gradient of the line a? 2 For each of the four curves, find the volume of revolution. In what way is the volume dependent on a? 3 Consider the function f (x) = ax n, where n = 1--2- , 1, 2 and 4 and a = 1. Find the area enclosed by the curve, the x-axis and the line x = 1 for each of the four values for n. In what manner is the area dependent on the value of n? 4 For each of the four curves, find the volume of revolution. In what way is the volume dependent on the value of n? 5 Finally, compare the sizes of the areas found in parts 1 and 3 to the volumes found in parts 2 and 4. In particular, investigate the ratio of the volume of V revolution V to the area A. In what way does the ratio ---- depend on the values A of a and n for the general function f (x) = ax n? What happens to the ratio as n → ∞ when a = 1? 6 Write a brief report detailing your findings being careful to illustrate your work with graphs and with calculations.

Chap 06 SM Page 270 Thursday, October 12, 2000 10:59 AM

270

Specialist Mathematics

summary Common antiderivatives • The table below lists common antiderivatives.

f(x)

F(x) ax n + 1

Ax n

--------------- + c n+1

1 --x

loge kx + c e kx ------- + c k – cos kx ------------------ + c k sin kx -------------- + c k tan kx -------------- + c k x Sin –1 --- + c a

e kx sin kx cos kx sec2kx 1 -------------------- , x ∈ (–a, a) 2 a – x2 –1 -------------------- , x ∈ (–a, a) a2 – x2 a ---------------a2 + x2

x Cos –1 --- + c a x Tan –1 --- + c a

Substitution where the derivative is present in the integrand • •

[ f ( x ) ]n + 1 f ′( x ) [ f ( x ) ] n dx = ------------------------- + c (n + 1)

∫ f ′( x ) - dx = log f ( x ) + c ∫ ----------f ( x) e

Linear substitution The integral

∫ f ( x) [g( x)]

n

dx, n ≠ 0 may be successfully antidifferentiated using

the substitution u = g(x), provided that g(x) is linear. The function f (x) must be written in terms of y also.

Useful trigonometric identities • Trigonometric identities are used to integrate even and odd powered trigonometric functions: sin2ax = 1--2- (1 − cos 2ax) cos2ax = 1--2- (1 + cos 2ax) sin ax cos ax =

1 --2

sin 2ax

Chap 06 SM Page 271 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

271

Antidifferentiation using partial fractions Many rational expressions can be antidifferentiated by transforming the expressions into partial fractions. Two common types are shown below.

Rational expression

Equivalent partial fraction

f ( x) ---------------------------------------( ax + b ) ( cx + d ) where f(x) is a linear function

A B --------------- + --------------ax + b cx + d

f ( x) ---------------------2( ax + b ) where f(x) is a linear function

B A ---------------------2- + --------------ax +b ( ax + b )

Definite integrals •

∫

b

f ( x ) dx = [ F ( x ) ] ab

a

= F(b) − F(a), where F(x) is an antiderivative of f(x).

• The definite integral

∫

b

f ( x ) dx can be found only if the integrand f(x) exists for

a

all values of x in the interval [a, b]; that is, a ≤ x ≤ b.

Areas under curves y

•

y = f (x)

Area =

∫

b

f ( x ) dx

a

0

a

b

x

y

•

a

Area =

b

∫ g( x ) dx a

x

0

b

y = g (x) y

•

y = f (x)

0 c

a

•

y b c

b

∫

c

∫

c

f ( x ) dx +

a

∫

b

f ( x ) dx

c

x

x = f (y)

0 a

Area =

x

Area =

a

f ( y ) dy +

b

∫ f( y ) dy c

Chap 06 SM Page 272 Thursday, October 12, 2000 10:59 AM

272

Specialist Mathematics

Areas between curves y

•

a

•

P

f (x)

0 c

g (x) x

b

Area =

[ g ( x ) – f ( x ) ] dx +

a

y

Area =

x = g (y) b

∫

c

b

∫ [ f ( x ) – g( x ) ] dx c

b

∫ [ g( y ) – f ( y ) ] dy a

x = f (y)

a x

0

Volumes of solids of revolution b

∫ [ f ( x ) ] dx . About y-axis: V = π [ f ( y ) ] dy ∫

• About x-axis: V = π

2

a

b

•

2

a

• Between two functions f (x) and g(x) where f (x) ≥ g(x): V = π

b

∫ [ f ( x)]

2

2

– [ g ( x ) ] dx

a

Approximate evaluation of definite integrals and areas • Approximate measure of

∫

b

f ( x ) dx using the midpoint rule:

a

x0 + x1 x1 + x2 xn – 1 + xn f ( x ) dx ≈ δx f  ---------------- + f  ---------------- + . . . + f  -----------------------       2 2 2 a where: n = the number of intervals used b–a δx = -----------n x0 = a xn = b

∫

b

• Approximate measure of

∫

b

f ( x ) dx using the trapezoidal rule:

a

δx f ( x ) dx ≈ ------ [ f ( x 0 ) + 2 f ( x 1 ) + 2 f ( x 2 ) + . . . + 2 f ( x n – 1 ) + f ( x n ) ] 2 a where: n = the number of intervals used b–a δx = -----------n x0 = a xn = b.

∫

b

Chap 06 SM Page 273 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

273

CHAPTER review Multiple choice 1 The expression A

∫u

5

∫ ( x – 1 )( x – 2x ) B ∫ u du 2

1 --2

du

5

dx is equal to:

6A

∫

5

C 2 u 5 du

D

∫ 5u

4

du

E

∫u

4

du

sin x - is: 2 An antiderivative of -----------cos 3 x

6A

–1 B -----------cos 2 x

1 A -----------cos 4 x 3 The expression

∫

A 2 u 4 du

–1 C ----------sin 2 x

1 D ----------------2 cos 2 x

∫ 6 sec 3x tan 3x dx is equal to: B u du C ∫ ∫ u du 2

1 E ----------------4 cos 2 x

4

1 --2

4

6A

4

D

∫u

2

∫

E 2 u 2 du

du

4 The antiderivative of x(x + 2)10 is: A x +c

B (x + 2) + c

( x + 2 ) 11 ( 12x – 11 ) D ------------------------------------------------ + c 132

E x(x + 2)11 + c

11

5

∫x

( x + 2 ) 11 ( 11x – 2 ) C --------------------------------------------- + c 132

11

2 – x dx is equal to: 5 ---

3 ---

5 ---

3 ---

6B 3 ---

2 --- ( 2 5

– x)2 – 2(2 – x)2 + c B

D

1 --- ( 2 5

2 - ( 2 – x ) 2 ( 3x + 4 ) + c – x ) 2 + 2 ( 2 – x ) 2 + c E – ----15

3

3

1

∫

D

∫

 u --2- + 2u --2- du  

5

3 ---

– x )2 – 3( 2 – x )2 + c C x2( 2 – x )2 + c ---

6 Using an appropriate substitution,  u --2- + u --2- du  

5 --- ( 2 2

5 ---

A

A

6B

∫e

2x

e x – 1 dx is equal to: 3

1

B

∫

 2u --2- + u --2- du  

E

∫

 u --2- + u --2- du  

3

5

1

C

6B 5  u --2-

∫

+

3 --2u 2

+

1 --u 2



du

Chap 06 SM Page 274 Thursday, October 12, 2000 10:59 AM

274 6C

Specialist Mathematics

7 Using an appropriate substitution, A D

6C

∫ (u ∫ (u

4

– u 2 ) du

B

3

– u 5 ) du

E

D

6D

6D

2

2

2

5

– u 3 ) du

4

B 2x − 1 + sin 2x

∫ ( u – u ) du ∫ ( –u – 1 ) du 2

4

B

4

10 The expression

C 2x + cos 2x

E 2x − sin 2x

∫ sin x dx is equal to: ∫ ( u – 2u ) du ∫ ( u – u + 1 ) du

9 Using the appropriate substitution, A

6C

3

π π 8 If f ′(x) = 4sin2x and f ---  = --- , then f (x) is equal to: 4  2 π A 4 cos 2 x + --- – 2 2 D 2x – cos 2x

6C

∫ cos x sin x dx is equal to: C (u ∫ u cos x dx ∫ ∫ ( u – u ) dx

E

5

4

4

2

C

∫ ( 2u

2

– 1 – u 4 ) du

2

∫ ( 2 + tan x ) dx is equal to: 2

A x + sec2x + c D x + tan x + c

B 2x + sec2x + c E xtan x + c

C tan x + c

1 1 1 1 = ----------- – -----------, x > 5 , an antiderivative of ----------------------------11 Given that ----------------------------is: 2 2 x – 5 x – 4 x – 9x + 20 x – 9x + 20 x–5 A log e  -----------  x – 4

x–4 B log e  -----------  x – 5

1 D log e ( x – 5 ) – ------------------2( x – 4)

1 1 E ------------------2- – ------------------2( x – 5) ( x – 4)

12 The expression

C loge(x2 − 9x + 20)

- dx is equal to: ∫ (-----------------x + 1) 2x + 3

2

–2 A ------------ + c x+1

–2 B ------------------2- + 3 log e ( x + 1 ) + c ( x + 1)

1 C log e ( x + 1 ) + ------------ + c x+1

1 D 2 log e ( x + 1 ) – ------------ + c x+1

E loge(x + 1)2 + c

6E

13 The integral

∫

b

a

A (–9, 9) D R

1 ------------------ dx can be evaluated over the largest domain of: 9 – x2 B [–3, 3] E (–3, 3)

C (–3, 0)

Chap 06 SM Page 275 Thursday, October 12, 2000 10:59 AM

Chapter 6

14 The value of

∫

1

6E 1 --3

A loge2

B

D loge4

E unable to be calculated.

The expression

275

x2 ------------- dx is: 3 x +1

–1

15

Integral calculus

∫

π

C 3 loge2

loge2

2x cos x 2 dx is equal to:

6E

0

A cos π

B sin (π 2)

C 0

D 1

E 2

16 The integral representing the shaded area of this curve is equal to: A 2×

∫

1

( x2

y

– 1 ) dx

6E

y = x2 – 1

0

B

∫

1

( x 2 – 1 ) dx

x

0

–1

C 2×

0

∫ (x

2

– 1 ) dx

1

1

D

∫ ( 1 – x ) dx ∫ ( x – 1 ) dx 2

0

2

E

2

0

17 The area between the curve y = sin x and the line y = x from x = 0 to x = 1 (see diagram) is approximately equal to: y

6F

y=x

1 0

A B C D E

1

–π 2

y = sin x x

0.04 square units 1.04 square units 0.54 square units 0.84 square units 0.34 square units

18 The shaded area (in square units) on the graph below is equal to: y

0

A

16 -----3

B 16

C

32 -----3

6F

y = (x – 2)2

4 x

D

8 --3

E 8

Chap 06 SM Page 276 Thursday, October 12, 2000 10:59 AM

276

Specialist Mathematics

Questions 19 and 20 refer to the shaded area in the figure below. y

y= x (1, 1) x

0 y=2–x

6G

2

19 The volume generated when the region is rotated about the x-axis is equal to: 1

∫ π ( 4 – 3x + x ) dx ∫ π ( 2 – x – x ) dx ∫

A π

0 1

C

2

0 2

E

1

∫ ( 4 + x – x ) dx π ( 4 – 4x + x + x ) dx ∫

B π

( 4 – 2x 2 + x 4 – x ) dx

4

0 1

4

D

2

4

0

2

0

6G

20 The volume generated when the region is rotated about the y-axis is equal to: 2

∫ π y dy ∫ π ( 2 – y – y ) dy ∫

A π

0 2

C E

( 2 – y ) dy + π

1 2

2

0 2

2

∫ π y ∫

B π

( 2 – y ) dy

D

1

∫y

2

dy

0

2

dy + π

1

1

∫ ( 2 – y ) dy 0

2

0

6H

21 The approximate value of A C E

6H

∫

4

ex ----- dx using the trapezoidal rule and 3 equal intervals is: x

1 1 2 3 4 ------ (6e + 3e + 2e + e ) 12 1 2 3 4 ------ (6e + 6e + 4e + e ) 12 1 2 3 4 ------ (12e + 12e + 8e + 3e ) 24

B D

1 --- (4e 8 1 --- (2e 4

+ 4e2 + 2e3 + e4) + 2e2 + e3 + 1--2- e4)

22 The approximate value of the area under the curve y = x2 + 1 from x = −1 to x = 1 (using the midpoint rule with four equal intervals) is: A 2.625 square units B 1.3125 square units C 2.5 square units D 2.75 square units E 1.95 square units

Short answer

6A

1 Find the antiderivative of:

6B 6C

2 Find the indefinite integral

( log e x ) 2 b -------------------x

a (cos x) esinx

∫

x ---------------- dx . x+1

3 Find: a

∫ cos 2x dx 2

b

∫ sin --4- cos --4- dx 2x

2x

Chap 06 SM Page 277 Thursday, October 12, 2000 10:59 AM

Chapter 6

Integral calculus

277

x 2 – 2x – 12 -. 4 Find an antiderivative of f (x) where f (x) = ---------------------------x 2 – 7x – 8

6D

5 Find f (x) if f ′(x) = sin 2x cos x and f (π) = 1.

6E

6 If f ′(x) = 2 1 – x 2 and f (0) = −3, find f (x). (Hint: Use the substitution x = sin θ to antidifferentiate.)

6E

7 Evaluate:

6E

a

∫

2

0

1 -------------2- dx 4+x

b

∫

1

–2

x ---------------- dx 2–x

8 a Sketch a graph which shows the region enclosed by the curve y = loge x, the line y = 2 and the x- and y-axes.

6E

b Find the area of this region. 9 What is the area bounded by the curve y = x2 + 2 and the line y = 5x − 4? 10 Find the volume generated when the area under the graph of y = e x, between x = −1 and x = 0, is rotated about the x-axis.

6F 6G

11 Find the volume of water in a hemispherical bowl of radius 8 cm if the depth is 3 cm.

6G

12 a Find an approximate value of

4

∫ 2x

2

dx using four equal intervals and:

0

i the midpoint rule ii the trapezoidal rule. b Which result is closest to the exact answer?

Analysis 1 a Find the area of the shaded region on the graph at right. b What is the volume generated when this region is rotated about the x-axis? c

If the region is rotated about the y-axis, find the approximate volume of the solid generated using the midpoint rule and four equal intervals. (Give your answer correct to 4 decimal places.) y

y = tan x

1 0

–π 4

–π 2

x

6H

Chap 06 SM Page 278 Thursday, October 12, 2000 10:59 AM

278

Specialist Mathematics

2 The side view of the right side of a wine glass vessel can be modelled by two curves which join at x = e: y

0 1 e

5 x

y = 2loge x, 0 < x ≤ e (red curve) y = x2 − 2ex + e2 + c, e ≤ x ≤ 5 (blue curve) (All measurements are in centimetres.) a Show that the value of c is 2 and find the height of the vessel correct to 2 decimal places. The vessel is formed when the region between the curves and the y-axis is rotated about the y-axis. b Find the volume of wine in the glass when the depth is 2 cm. c What is the maximum volume of wine that the glass can hold (using maximum height to the nearest mm)? 3 A below-ground skating ramp is to be modelled by the curve 2 y = ---------------------, – 5.98 ≤ x ≤ 5.98 . 36 – x 2 y Ground 4.086 Level

–6

CHAPTER

test yyourself ourself

6

0

6 x

This is shown above, where the line y = 4.086 represents ground level. (All measurements are in metres.) (Give all answers correct to 2 decimal places.) a Find the maximum depth of the ramp. b Find the area under the curve. c Find the volume generated if this area is rotated about the x-axis. d If the ramp is 20 metres long, what is the volume of dirt which must be removed?

Integration

9 VCE co covverage Area of study Units 3 & 4 • Calculus

In this cha chapter pter 9A Revision of antidifferentiation 9B Integration of e x, sin x and cos x 9C Integration by recognition 9D Approximating areas enclosed by functions 9E The fundamental theorem of integral calculus 9F Signed areas 9G Further areas 9H Areas between two curves 9I Further applications of integration

344

Mathematical Methods Units 3 and 4

Revision of antidifferentiation As we have seen, the process of differentiation enables us to find the gradient of a function. The reverse process, antidifferentiation (or integration), will find the function for a particular gradient. Integration has wider applications including calculation of areas, volumes, energy, probability and many more quantities in science and business. d d Note that ------ f ( x) means differentiate f ( x) with respect to x; that is, ------ f ( x ) = f ′ ( x ) . dx dx So f ( x) is the antiderivative of f ′( x), denoted as f ( x) = where

∫ f ′ ( x ) dx

∫ means antidifferentiate, or integrate, or find an indefinite integral and dx indi-

cates that the integration of the function is with respect to x.

d ------ ( ax + c ) = a, where a and c are constants dx

Since

∫ a dx = ax + c

then

d ax n + 1 ------  --------------- = ax n dx  n + 1 

Since

∫

then

a xn + 1 a x n d x = ---------------- + c, n ≠ – 1 n+1

Note: The addition of the constant, c, is required when finding general antiderivatives. However, if we have to find an antiderivative, the c is to be allocated an actual number and for convenience the number chosen is zero. That is, an antiderivative means ‘let c = 0’, or ‘do not add on the c’. 2

3

2

For example, the antiderivative of 3x + 4x + 5 is x + 2x + 5x + c . An antideriva2

3

2

tive of 3x + 4x + 5 is x + 2x + 5x .

Properties of integrals d Since ------ is a linear operator, so too is its inverse, . Therefore, dx

∫

∫ [ f ( x) ± g( x)] d x = ∫ f ( x) d x ± ∫ g( x) d x That is, each term can be integrated separately, and

∫ k f( x) d x = k∫ f( x) d x That is, a ‘constant’ factor of the function can be taken to the front of the integral.

Chapter 9 Integration

345

WORKED Example 1 Antidifferentiate each of the following, giving answers with positive indices. 3 a 2x7 b 4x−3 c ------x THINK

WRITE

a

a

b

c

1

Integrate by rule; that is, add 1 to the index and divide by the new index.

2

Simplify.

1

Integrate by rule.

2

Simplify.

3

Express the answer with a positive index.

1

When a square root is involved, replace it with a fractional index.

2

Bring the x to the numerator and change the sign of the index.

∫

2x 8 2x 7 dx = -------- + c 8 x8 = ----- + c 4 4x –2 4x –3 dx = ---------- + c –2 = −2x−2 + c 2 = − ----2x

b

∫

c

∫ ------x- dx = ∫ ----x- dx

3

3

1 --2

∫

=3 x

– 1--2-

dx

1 ---

3

3x 2 -+c = ------1

Integrate by rule.

--2

1 ---

4

Simplify.

= 6x 2 + c

5

Write the answer in the form it was given.

= 6 x+c

WORKED Example 2

Find the following indefinite integral. ( x – 1)(3 x + 5) d x

∫

THINK 1

Expand the expression.

2

Collect like terms.

3

Integrate each term separately.

4

Simplify each term.

WRITE

∫ ( x – 1 ) ( 3x + 5 ) dx = ∫ ( 3x 2 – 3x + 5x – 5 ) dx = ∫ ( 3x 2 + 2x – 5 ) dx 3x 3 2x 2 = -------- + -------- – 5x + c 3 2 = x 3 + x 2 – 5x + c

346

Mathematical Methods Units 3 and 4

Integration of (ax + b)n By applying the chain rule for differentiation: d ------ ( ax + b ) n + 1 = a ( n + 1 ) ( ax + b ) n dx so a ( n + 1 ) ( ax + b ) n dx = ( ax + b ) n + 1 + c

∫

∫ ∫ ( ax + b )n dx =

or

a ( n + 1 ) ( ax + b ) n dx = ( ax + b ) n + 1 + c

or

( ax + b ) n + 1 ----------------------------- + c a(n + 1)

WORKED Example 3

Antidifferentiate 4(5x − 2)3 by rule. THINK 1

2 3

WRITE

∫ 4 ( 5x – 2 )3 dx = 4 ∫ ( 5x – 2 )3 dx

Express as an integral and take 4 out as a factor.

4 ( 5x – 2 ) 4 = ------------------------- + c 5(4)

Apply the rule where a = 5 and n = 3.

= 1--5- (5x − 2)4 + c

Simplify the antiderivative by cancelling the fraction.

Integration of --1xSince then or

d 1 ------ log e x = --xdx 1 --- dx = log e x + c, where x > 0 x

∫

∫ x –1 dx = loge x + c .

WORKED Example 4 4 Antidifferentiate ------ . 7x THINK 1

Take

WRITE 4 --7

out as a factor.

- dx =  --- × --- dx ∫ ----∫  7 x 7x 4

4

= 2

Integrate by rule.

4 --7

1

1

∫ --x- dx

= 4--7- log e x + c

Chapter 9 Integration

Integration of (ax + b)–1 By applying the chain rule for differentiation: d a ------ log e ( ax + b ) = --------------- , dx ax + b where a and b are constants a 1 1 d --- × ------ log e ( ax + b ) = --------------- × --and ax + b a a dx

∫

So

1 = --------------ax + b 1 1 --------------- d x = --- log e ( ax + b ) + c ax + b a 1

∫ ( ax + b )–1 d x = --a- loge ( ax + b ) + c

or

1 Note that the a in the fraction --- is the derivative of the linear function ax + b. a 1 1 ----------------------------dx = ------------------ log e ( ax 2 + bx + c ) + c 1 2ax + b ax 2 + bx + c

∫

WORKED Example 5 5 Antidifferentiate --------------- . 2x + 3 THINK 1

2

Express as an integral and take 5 out as a factor.

WRITE 5

1

- dx = 5 --------------- dx ∫ -------------∫ 2x + 3 2x + 3 5 = --- log e ( 2x + 3 ) + c 2

Integrate by rule where a = 2.

WORKED Example 6

6x + 5 Find --------------dx . x2 THINK

∫

WRITE 6x + 5

- dx =  -----2- + ----2- dx ∫ -------------∫ x x  x2 6x

5

1

Express as separate fractions.

2

Simplify each fraction.

= ( 6x –1 + 5x –2 ) dx

3

Integrate each term separately by rule.

5x –1 = 6 log e x + ---------- + c –1

4

Simplify leaving the answer with positive indices.

∫

= 6 log e x – 5x –1 + c 5 = 6 log e x – --- + c x

347

348

Mathematical Methods Units 3 and 4

WORKED Example 7

Find the equation of the curve g(x) given that g′(x) = 3 x + 2 and the curve passes through (1, 2). THINK

WRITE

1

Write the rule for g′(x).

g′(x) = 3 x + 2

2

Rewrite g′(x) in index form.

g′(x) = 3x 2 + 2

3

Express g(x) in integral notation.

g(x) = ( 3x 2 + 2 ) dx

4

Antidifferentiate to obtain a general rule for g(x).

= 3x 2 ÷ 3--2- + 2x + c

Simplify.

3x 2 = -------- × 2--3- + 2x + c 1

1 ---

∫

1 ---

3 ---

3 ---

5

3 ---

g(x) = 2x 2 + 2x + c 7

Substitute coordinates of the given point into g(x). Find the constant of antidifferentiation, c.

8

State the rule for g(x) in the form that it is given.

6

As g(1) = 2,

3 ---

2(1)2 + 2(1) + c = 2 2+2+c =2 so c = −2 3 ---

g(x) = 2x 2 + 2x – 2 = 2 x 3 + 2x – 2

WORKED Example 8

If a curve has a stationary point (2, 3), and a gradient of 2x − k, where k is a constant, find: a the value of k b y when x = 1. THINK a

1 2

3

b

1

WRITE dy dy The gradient is ------ so write the rule for the gradient. a ------ = 2x – k dx dx dy Let ------ = 0 (as stationary points occur when the For stationary points, dx dy derivative is zero) and substitute the value of x into ------ = 0 , so 2x − k = 0 dx this equation. 2(2) − k = 0 as x = 2 Solve for k. 4 − k = 0 so k = 4 dy Rewrite the rule for the gradient function, using the b ------ = 2x – 4 dx value of k found in a above.

2

Integrate to obtain the general rule for y.

3

Substitute the coordinates of the given point on the curve to find the value of c.

4

State the rule for y. Substitute the given value of x and calculate y.

5

∫

y = ( 2x – 4 ) dx = x 2 – 4x + c Since curve passes through (2, 3), 3 = 22 − 4(2) + c 3=4−8+c c=7 So y = x2 − 4x + 7 When x = 1, y = (1)2 − 4(1) + 7 =4

349

Chapter 9 Integration

remember remember d 1. ------ f ( x ) = f ′ ( x ) dx 2. f ( x ) =

∫ f ′ ( x ) dx

∫ a dx = ax + c ax n + 1 4. ∫ ax n dx = --------------- + c , n ≠ −1 n+1 5. ∫ [ f ( x ) ± g ( x ) ] dx = ∫ f ( x ) dx ± ∫ g ( x ) dx 6. ∫ k f ( x ) dx = k ∫ f ( x ) dx ( ax + b ) n + 1 7. ∫ ( ax + b ) n dx = ----------------------------- + c , n ≠ −1 a(n + 1) 3.

8.

1

∫ --x- dx = loge x + c , where x > 0 or ∫ x –1 dx = loge x + c

1 1 --- log e ( ax + b ) + c or ( ax + b ) –1 dx = --- log e ( ax + b ) + c a a 10. Write the answer in the same form as the question unless otherwise stated. 9.

1

- dx = ∫ -------------ax + b

∫

9A WORKED

1

1 Antidifferentiate each of the following, giving answers with positive indices. a x c x7 d 3x5 e 5x−2 b x4 f

−2x4

g −6x−4

h 2 x

k

x –4 ------3

l

m x3

5 p ----3x WORKED

2

9 q ----2x

2 ---

r

x4 ----5

x3 ----2

n 4x --4-

o

x

8 ------x

t

–6 --------------( x x)

3

– 10 -------x6

s

Antidifferentiation

– 3--7-

2 Find the following indefinite integrals. a c e g i

∫ ( 2x + 5 ) dx ∫ ( 10x 4 + 6x 3 + 2 ) dx ∫ ( x 3 + 12 – x 2 ) dx ∫ 5 ( x 2 + 2x – 1 ) dx ∫ x ( x – 1 ) ( x + 4 ) dx

b d f h

∫ + 4x – 10 ) dx ∫ ( –4x 5 + x 3 – 6x 2 + 2x ) dx ∫ ( x + 3 ) ( x – 7 ) dx ∫ ( x 2 + 4 ) ( x – 7 ) dx ( 3x 2

Math

cad

Example

x

j

i

Math

cad

Example

Revision of antidifferentiation

Integrator

350

Mathematical Methods Units 3 and 4

3 multiple choice

∫ ( x 2 + x + 2 ) dx is equal to: A ∫ x 2 dx + x + 2 C ∫ ( x 2 + x ) dx + 2 E x 2 + x + ∫ 2 dx

B D

∫ x 2 dx + ∫ x dx + ∫ 2 dx x 2 + ∫ ( x + 2 ) dx

4 multiple choice

∫ x ( x + 3 ) dx is equal to: A ∫ x dx ∫ ( x + 3 ) dx C ( x + 1 ) ∫ x dx E ∫ ( x 2 + 3x ) dx WORKED

Example

3

∫

B x ( x + 3 ) dx D

∫ x dx + ∫ ( x + 3 ) dx

5 Antidifferentiate each of the following by rule. a (x + 3)2 b (x − 5)3 c 2(2x + 1)4 e (6x + 5)4 f 3(4x − 1)2 g (4 − x)3 i 4(8 − 3x)4 j −3(8 − 9x)10 k (2x + 3)−2 m 6(4x − 7)−4 n (3x − 8)−6 o (6 − 5x)−3

d h l p

−2(3x − 4)5 (7 − x)4 (6x + 5)−3 −10(7 − 5x)−4

6 multiple choice

∫ 3 ( x + 2 )4 dx is equal to: A 3 + ∫ ( x + 2 ) 4 dx C 3 ∫ ( x + 2 ) 4 dx E ( x + 2 ) 4 ∫ 3 dx WORKED

Example

4

WORKED

Example

5

7 Antidifferentiate the following. 3 a --- dx b x 7 ------ dx d e 3x 3 ------------ dx g h x+3 4 --------------- dx j k 3x + 2 –5 --------------- dx m n 3 + 2x 3 ------------------ dx p q 6 – 11x

∫

B D

8

∫ --x- dx

∫ 3 dx + ∫ ( x + 2 )4 dx 4 3 ∫ dx × ∫ ( x + 2 ) dx

6

c

- dx ∫ ----5x - dx ∫ ----------x+3

∫

- dx ∫ ----7x

4

f

∫

- dx ∫ ----------x+4

–2

i

- dx ∫ ----------x+5

∫

- dx ∫ -------------5x + 6

8

l

--------------- dx ∫ 2x –5

–2

o

dx ∫ ----------5–x

–2

r

- dx ∫ 5-------------– 2x

∫

- dx ∫ -------------6 + 7x

∫

- dx ∫ 4-------------– 3x

1

–6 3

1

–8

Chapter 9 Integration

8 multiple choice 6 ------------ dx is equal to: x+5 1 A 6 ------------ dx x+5

351

∫

∫

C

B

1

- dx ∫ 6dx + ∫ ----------x+5

D

6 E --------------------------( x + 5 ) dx

1

- dx ∫ 6dx ∫ ----------x+5 ∫ 6dx --------------------------∫ ( x + 5 ) dx

∫

Example

6

( 2x + 7 ) 9 Find -------------------- dx . x

∫

10 For the following mixed sets, find:  x 4 + 2x + 1--- ( 3x + 1 ) 5 dx b a  x 3 –5 --------------- dx ------------------ dx d e 2x + 1 6 – 10x ( x + 4 )2 3 ( 4x + 1 ) –3 dx g ------------------- dx h 2x 3 2 --– 1-- x + ----------- dx j ( 5x 2 – 2x + 3x 3 ) dx k   3–x m WORKED

Example

7

∫ ∫ ∫

∫

c

∫

f

∫

i

∫

∫

l

x2

+ 2x – 1

dx ∫ -------------------------x

n

∫

3x 2 + 2x – 1 ----------------------------- dx x2

∫

∫ 4 ( 2x – 5 )5 dx ( x – 5)( x + 3) - dx ∫ --------------------------------x3 x2 + x4

dx ∫ ---------------x

10 – x + 2x 4 ----------------------------dx x3

11 Find the equation of the curve f (x) given that: a f ′(x) = 4x + 1 and the curve passes through (0, 2) b f ′(x) = 5 − 2x and the curve passes through (1, −1) c f ′(x) = x−2 + 3 and the curve passes through (1, 4) d f ′(x) = x + x and f (4) = 10 1 ---

e f ′(x) = x 3 – 3x 2 + 50 and f(8) = −100 1 f f ′(x) = ------- – 2x and f (1) = −5 x g f ′(x) = (x + 4)3 and the curve passes through (−2, 5) h f ′(x) = 8(1 − 2x)−5 and f(1) = 3 i f ′(x) = (x + 5)−1 and the curve passes through (−4, 2) 8 j f ′(x) = --------------- and f(3) = 7 7 – 2x WORKED

Example

8

12 If a curve has a stationary point (1, 5), and a gradient of 8x + k, where k is a constant, find: a the value of k b y when x = −2. kx + x - , where k is a constant, and a stationary point 13 A curve g(x) has g′ ( x ) = -----------------x2 (1, 2). Find: a the value of k b g(x) c g(4).

HEET

SkillS

WORKED

9.1

352

Mathematical Methods Units 3 and 4

Integration of ex, sin x and cos x Integration of the exponential function ex Since

d ------ ( e x ) = e x dx

then

∫ e dx = e

and or

x

x

+c

d ------ ( e kx ) = ke kx , where k is a constant dx d 1 1 ------  --- × e kx = --- × ke kx  dx  k k = e kx

1 Therefore, e kx dx = --- e kx + c. k

∫

WORKED Example 9 Antidifferentiate each of the following. a 3e4x

e –5 x b ---------4

c (e x − 1)2

THINK a

b

1

Integrate by rule where k = 4.

2

Simplify.

1

Rewrite the function to be integrated so that the coefficient of the e term is clear.

WRITE a

∫

3e 4 x 3e 4 x dx = ---------- + c 4 = 3--4- e 4 x + c

b

∫

e –5 x ---------dx = 4

∫ --4- e–5 x dx 1

2

Integrate by rule where k = −5.

1 –5 x --- e 4 = ------------ + c –5

3

Simplify the antiderivative.

1 -+c = 1--4- e –5 x × ----–5 1 –5 x -e = – ----+c 20

c

1

Expand the function to be integrated.

2

Integrate each term by the rule.

c

∫ ( e x – 1 )2 dx = ∫ ( e2 x – 2e x + 1 ) dx = 1--2- e 2 x – 2e x + x + c

Chapter 9 Integration

Integration of trigonometric functions d ------ (sin ax) = a cos ax dx

Since it follows that

d ------ (cos ax) = −a sin ax dx

and 1

∫ sin ax dx = --a- cos ax + c 1 ∫ cos ax dx = --a- sin ax + c

WORKED Example 10 Antidifferentiate the following. x a sin 6x b 8 cos 4x c 3 sin  – ---  2 THINK

WRITE

a Integrate by rule.

a

∫ sin 6x dx = – --6- cos 6x + c

b

b

∫ 8 cos 4x dx = --4- sin 4x + c

c

1

Integrate by rule.

2

Simplify the result.

1

Integrate by rule.

2

Simplify the result.

1

8

= 2 sin 4x + c

c

∫ 3 sin  – --2- dx = -----–1--2-- cos  – --2- x

–3

x = 6 cos  – --- + c  2

WORKED Example 11 ∫

Find ( 2e 4 x – 5 sin 2 x + 4 x ) d x˙. THINK 1

WRITE

∫ ( 2e4 x – 5 sin 2x + 4x ) dx

Integrate each term separately.

5 4 - cos 2x + --- x 2 + c = 2--4- e 4 x – –----2 2

2

= 1--2- e 4 x + 5--2- cos 2x + 2x 2 + c

Simplify each term where appropriate.

remember remember 1. 2.

1

∫ e dx = e + c and ∫ e dx = --k- e + c 1 1 ∫ sin ax dx = − --a- cos ax + c and ∫ cos ax dx = --a- sin ax + c x

x

x

kx

kx

353

354

Mathematical Methods Units 3 and 4

9B WORKED

Example

d hca

Mat

9 Integrator

Integration of ex, sin x and cos x

1 Antidifferentiate each of the following. b e4x a e2x −3x d e e 5e5x 2e 3 x e6 x g ------h ---------2 3 j

−8e−2x

x

k e --3-

e−x 7e4x

i

−3e6x

l

0.1e 4

x ---

o e x + e−x

–x

x

m 3e --2e x – e–x p -----------------2

c f

n 3e -----3-

2 Find an antiderivative of (1 + e x)2. 3 Find an antiderivative of (e x − 1)3. 4 Find an antiderivative of x3 − 3x2 + 6e3x. 5 multiple choice If f ′(x) = e2x + k and f(x) has a stationary point (0, 2), where k is a constant, then: a k is equal to: C 1 D −1 E −e A e B e2 b f(1) is equal to: A e2 − 1 WORKED

Example

10

B e2 +

1 --2

C

1 2 --- e 2

6 Antidifferentiate the following. a sin 3x b sin 4x cos 2x e sin (−2x) d ---------------3 4 sin 6x h 8 cos 4x g ------------------3 x j −2 cos (−x) k sin --3

+

1 --2

E 1 1--2-

D e4

c

cos 7x

f

cos (−3x)

i

−6 sin 3x

l

x cos --2

x m 3 sin  – ---  4

x n −2 sin --5

x o 4 cos --4

x p −6 cos  – ---  2

2x q 4 sin -----3

r

s

5x −2 sin -----2

πx v 3 cos -----2

t

7x −3 cos -----4

πx w −2 cos -----3

3x 6 cos -----4

u 5 sin π x 4x x −sin  – ------  π

Chapter 9 Integration

WORKED

Example

11

355

7 Find: a

∫ ( sin x + cos x ) dx

b

∫ ( sin 2x – cos x ) dx

c

∫ ( cos 4x + sin 2x ) dx

d

∫  sin --2- – cos 2x dx

e

∫ ( 4 cos 4x – --3- sin 2x ) dx

f

∫ ( 5x + 2 sin x ) dx

g

∫  3 sin -----2- + 2 cos -----3- dx

h

∫ ( 3e6 x – 4 sin 8x + 7 ) dx

1

πx

πx

x

8 Find the antiderivative of e4x + sin 2x + x3. 9 Find an antiderivative of 3x2 − 2 cos 2x + 6e3x. 10 Antidifferentiate each of the following. 1 a x 3 – --------------- + e 2 x 2x + 3 c

x --x sin --- + e 2 – ( 3x – 1 ) 4 3

–x -----x x e 3 sin --- – 2 cos --- – e 5 2 3

b x2 + 4 cos 2x − e−x x 1 d --------------- + e 4 x + cos --5 3x – 2 f

πx x + 2x – 2 sin ------ + 5 3

11 In each of the following find f(x) if:

Math

cad

π a f ′(x) = cos x and f  --- = 5  2

Antidifferentiation

b f ′(x) = 4 sin 2x and f(0) = −1 x f ′(x) = 3 cos --- and f(π) = 9 2 4 x x d f ′(x) = cos --- − sin --- and f(2π) = −2. 4 2 c

πx dy 12 If ------ = sin ------ + k , where k is a constant, and y has a stationary point (3, 4), find: 6 dx a the value of k b the equation of the curve c

y when x = 6.

13 A curve has a gradient function f ′(x) = 4 cos 2x + ke x, where k is a constant, and a stationary point (0, −1). Find: a the value of k b the equation of the curve f(x) c

π f  --- correct to 2 decimal places.  6

356

Mathematical Methods Units 3 and 4

Integration by recognition d As we have seen, if ------ [f(x)] = g ( x ) dx then

∫ g ( x ) d x = f ( x ) + c , where g(x) = f ′(x).

This result can be used to determine integrals of functions, which are too difficult to antidifferentiate, via differentiation of a related function.

WORKED Example 12

a Find the derivative of the function y = (5x + 1)3. b Use this result to deduce the antiderivative of 3(5x + 1)2. THINK a

1 2 3 4 5 6 7 8

b

1

2

3

WRITE

Write the function and recognise that the a y = (5x + 1)3 chain rule can be used. Let u equal the function inside the brackets. Let u = 5x + 1 du du ------ = 5 Find ------ . dx dx Express y in terms of u. y = u3 dy dy ------ = 3u 2 Find ------ . du du dy dy du ------ = ------ × -----Write the chain rule. dx du dx dy dy ------ = 3u 2 × 5 Find ------ using the chain rule. dx dx Replace u with the expression inside the = 15(5x + 1)2 brackets and simplify where applicable. dy Since ------ dx = y + c 1, express the dx relationship in integral notation. dy Remove a factor from ------ so that it dx resembles the integral required.

∫

Divide both sides by the factor in order to obtain the integral required.

b

∫ 15 ( 5x + 1 )2 dx = ( 5x + 1 )3 + c1 ∫

5 3 ( 5x + 1 ) 2 dx = ( 5x + 1 ) 3 + c 1

∫ 3 ( 5x + 1 )2 dx = --5- ( 5x + 1 )3 + c , where 1

c1 c = ----5 Therefore, the antiderivative of 3 ( 5x + 1 ) 2 is 1--5- ( 5x + 1 ) 3 + c.

Note that the shorter form of the chain rule below can be used to differentiate. If f ( x ) = [ g ( x ) ] n then f′ ( x ) = ng′ ( x ) [ g ( x ) ] n – 1

Chapter 9 Integration

WORKED Example 13 3

a Differentiate e x .

3

b Hence, antidifferentiate 6x2 e x .

THINK

WRITE

a

Write the equation and apply the chain rule to differentiate y. Let u equal the index of e. du Find ------ . dx Express y in terms of u. dy Find ------ . du dy Find ------ using the chain rule dx and replace u.

a

dy Express ------ in integral notation. dx

b

1 2 3 4 5 6

b

1 2

357

y = ex

3

Let u = x3 du ------ = 3x 2 dx y = eu dy ------ = e u du dy ------ = 3x 2 e u , applying the chain rule dx 3 = 3x 2 e x

Multiply both sides by a constant to obtain the integral required.

∫ 3x 2 e x dx = e x + c1 2 ∫ 3x 2 e x dx = 2e x + 2c 1 ∫ 6x 2 e x dx = 2e x + c , where c = 2c1. 3

3

3

3

3

3

Therefore, the antiderivative of 6x 2 e x is 2e x + c . 3

3

Note that the shorter form of the chain rule below can be used to differentiate. dy If y = e f ( x ) then ------ = f ′ ( x )e f ( x ) dx

WORKED Example 14

a Find the derivative of sin (2x + 1) and use this result to deduce the antiderivative of 8 cos (2x + 1). x -. b Differentiate loge (5x2 − 2) and hence antidifferentiate ----------------2 5x – 2 THINK WRITE a 1 Let f(x) equal the rule. a f(x) = sin (2x + 1) f ′(x) = 2 cos (2x + 1) 2 Differentiate using f ′(x) = g′(x) cos [g(x)] where f(x) = sin [g(x)]. 3

Express f(x) using integral notation.

4

Multiply both sides by whatever is necessary for it to resemble the integral required.

5

Write the integral in the form in which the question is asked.

∫ 2 cos ( 2x + 1 ) dx = sin ( 2x + 1 ) + c1 4 ∫ 2 cos ( 2x + 1 ) dx = 4 sin ( 2x + 1 ) + c ∫ 8 cos ( 2x + 1 ) dx = 4 sin ( 2x + 1 ) + c Continued over page

358

Mathematical Methods Units 3 and 4

THINK

WRITE

b

b

1 2

Let f(x) equal the rule. g′ ( x ) Differentiate using f ′ ( x ) = -----------g( x) where f(x) = loge [g(x)].

3

Express f(x) using integral notation.

4

Take out a factor so that f ′(x) resembles the integral required.

5

f(x) = loge (5x2 − 2) 10x f ′ ( x ) = ---------------5x 2 – 2 10x

---------------- dx = log e ( 5x 2 – 2 ) + c 1 2–2 ∫ 5x x - dx = log e ( 5x 2 – 2 ) + c 1 10 ---------------5x 2 – 2

∫

x

- log e ( 5x 2 – 2 ) + c ---------------- dx = ----10 2–2 ∫ 5x

Divide both sides by the factor to obtain the required integral.

1

The antiderivative of x ----------------- is 5x 2 – 2

1 -----10

log e ( 5x 2 – 2 ) + c

WORKED Example 15 Differentiate x cos x and hence find an antiderivative of x sin x. THINK 1 Write the rule. 2

Apply the product rule to differentiate x cos x.

3

Express the result in integral notation. (Do not add c, as an antiderivative is required.)

4

Express the integral as two separate integrals.

5

Simplify by integrating. (Do not add c.)

6

Make the expression to be integrated the subject of the equation.

7

Simplify.

WRITE Let y = x cos x dy ------ = x (−sin x) + (cos x)(1) dx = −x sin x + cos x = cos x − x sin x

∫

∴ ( cos x – x sin x )dx = x cos x

∫

∫

() cos x dx – x sin x dx = x cos x

∫ – ∫ x sin x dx =

sin x – x sin x dx = x cos x x cos x – sin x

∫ x sin x dx = sin x – x cos x Therefore, an antiderivative of x sin x is sin x − x cos x.

Chapter 9 Integration

359

WORKED Example 16 5x + 1 4 a Show that --------------- = 5 – ------------ . x+1 x+1

5x + 1 b Hence, find --------------- d x . x+1

∫

THINK

WRITE

a Use algebraic long division to divide the numerator into the denominator.

5 a x + 1 ) 5x + 1 −(5x + 5) –4

b

b

1

Write the expression using integral notation.

2

Express as two separate integrals.

3

Antidifferentiate each part.

5x + 1 4 So --------------- = 5 – -----------x+1 x+1

5x + 1

- dx =  5 – ------------ ∫ -------------∫  x + 1 x+1 4

4 = 5 dx – ------------ dx x+1 = 5x − 4 loge (x + 1) + c

∫

∫

cad

Math

The Mathcad file shown below may be used to check derivatives.

remember remember 1. To differentiate using the chain rule, use one of the following rules. (a) If f(x) = [g(x)]n then f ′(x) = ng′(x)[g(x)]n − 1 dy (b) If y = e f(x), ------ = f ′ ( x )e f ( x ) dx dy dy du (c) ------ = ------ × -----dx du dx (d) f ′(x) = g′(x) cos [g(x)] where f(x) = sin [g(x)] f ′(x) = −g′(x) sin [g(x)] where f(x) = cos [g(x)] g′( x) (e) f ′ ( x ) = ------------ where f (x) = loge[g(x)]. g( x)

∫

2. To antidifferentiate use g ( x ) dx = f ( x ) + c where g(x) = f ′(x).

Differentiator

360

Mathematical Methods Units 3 and 4

9C WORKED

Example

12

Mat

d hca

Differentiator

Integration by recognition

1 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. ii 12(3x − 2)7 a i (3x − 2)8 2 5 b i (x + 1) ii 5x(x2 + 1)4 1 ii ------------------c i 2x – 5 2x – 5 2 d i 4x + 3 ii ------------------4x + 3 e i (x2 + 3x − 7)4 ii (2x + 3)(x2 + 3x − 7)3 4x 1 ii -------------------f i ------------2 2 ( x – 1 )2 x –1 2 multiple choice The derivative of (x + 7)4 is 4(x + 7)3. a Therefore, the antiderivative of 4(x + 7)3 is: A (x + 7)4 + c B 1--4- (x + 7)4 + c 4 D 3(x + 7) + c E 12(x + 7)4 + c b The antiderivative of (x + 7)3 is: B 1--4- (x + 7)4 + c A (x + 7)4 + c 4 D 3(x + 7) + c E 12(x + 7)4 + c

C 4(x + 7)4 + c C 4(x + 7)4 + c

3 multiple choice

∫

If the derivative of (2x − 3)6 is 12(2x − 3)5, then 6 ( 2x – 3 ) 5 dx is: A 2(2x − 3)6 + c D 6(2x − 3)6 + c WORKED

Example

13

WORKED

Example

14

B 4(2x − 3)6 + c E 1--2- (2x − 3)6 + c

C (2x − 3)6 + c

4 For each of the following differentiate i and hence antidifferentiate ii. ii 2e4x − 5 a i e4x − 5 6 − 5x b i e ii 10e6 − 5x 2 2 c i ex ii x e x 2 2 d i ex – x ii ( 1 – 2x )e x – x 5 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i sin (x − 5) ii cos (x − 5) b i sin (3x + 2) ii 6 cos (3x + 2) c i cos (4x − 7) ii sin (4x − 7) d i cos (6x − 3) ii 3 sin (6x − 3) e i sin (2 − 5x) ii 10 cos (2 − 5x) f i cos (3 − 4x) ii −2 sin (3 − 4x) 20 g i loge (5x + 2) ii --------------5x + 2 12x ii ------------h i loge (x2 + 3) x2 + 3 x–2 ii ---------------i i loge (x2 − 4x) x 2 – 4x

Chapter 9 Integration

WORKED

Example

15

361

6 Differentiate i and hence find an antiderivative of ii. a i x cos x + 2 sin x ii x sin x 2 ( x cos x – sin x ) sin x ii -----------------------------------------b i ----------x x2 c i e x sin x ii 3e x (sin x + cos x) d i x sin x ii x cos x x e i xe ii x e x 7 For each of the following differentiate i and use this result to antidifferentiate ii. a i (2x − 3x2)6 b i

Example

x 3 + 2x

3x – 2 1 8 a Show that --------------- = 3 + ----------- . x–1 x–1

b

– 2- dx . -------------Hence, find 3x x–1

5x + 8 2 9 a Show that --------------- = 5 − ------------ . x+2 x+2

b

+ 8- dx . -------------Hence, find 5x x+2

8x – 7 5 10 a Show that --------------- = 4 + --------------- . 2x – 3 2x – 3

b

– 7- dx . -------------Hence, find 8x 2x – 3

6x – 5 4 11 a Show that --------------- = −3 + --------------- . 3 – 2x 3 – 2x

b

– 5- dx . -------------Hence, find 6x 3 – 2x

b

hence find tan x dx .

∫

16

∫ ∫ ∫

12 If y = loge (cos x): dy a find -----dx

∫

cos x 1 -. 13 Differentiate ------------ and hence find an antiderivative of ----------sin x sin 2 x x -. 14 Differentiate loge (3x2 − 4) and hence find an antiderivative of ---------------3x 2 – 4 15 Differentiate sin (ax + b) and hence find an antiderivative of cos (ax + b). (Here, a and b are constants.) 16 Differentiate cos (ax + b) and hence find an antiderivative of sin (ax + b). (Here, a and b are constants.) 17 Differentiate eax constants.)

+ b

and hence find an antiderivative of eax

+ b

. (Here, a and b are

18 Antidifferentiate each of the following. a sin (3π x + 1)

b cos (1 − 4π x)

c

eπ x + 3

πx d sin  2 + ------  3

πx e 3 cos  ------ + 5 2 

f

cos x esin x

ET SHE

Work

WORKED

ii 6x5(1 − 3x)(2 − 3x)5 3x 2 + 2 ii ---------------------x 3 + 2x

9.1

362

Mathematical Methods Units 3 and 4

Approximating areas enclosed by functions There are several ways of finding an approximation to the area between a graph and the x-axis. We will look at three methods: 1. the lower rectangle method 2. the upper rectangle method 3. the trapezoidal method. y

The lower rectangle method

f(x)

Consider the area between the curve f(x) shown at right, the x-axis and the lines x = 1 and x = 5. If the area is approximated by ‘lower’ rectangles of width 1 unit then the top of each rectangle is below the R1 R 2 R 3 R 4 graph but touches the curve at one point. (In this case the left-hand corner of the rectangle touches the graph.) 0 1 2 3 4 5 x So, the height of rectangle R1 is f(1) units and the area of R1 = 1 × f(1) square units (area of a rectangle = height × width). Similarly, the area of R2 = 1 × f(2) square units, the area of R3 = 1 × f(3) square units, the area of R4 = 1 × f(4) square units. Therefore, the approximate area under the graph between the curve f(x), the x-axis and the lines x = 1 to x = 5 is 1[f(1) + f(2) + f(3) + f(4)] square units, (the sum of the area of the four rectangles). If the same area was approximated using rectangle widths of 0.5 there would be 8 rectangles and the sum of their areas would be: 0.5[f(1) + f(1.5) + f(2) + f(2.5) + f(3) + f(3.5) + f(4) + f(4.5)] square units. From the diagram it can be seen that the lower rectangle approximation is less than the actual area.

WORKED Example 17

y

f(x)

Find an approximation for the area between the curve f(x) shown and the x-axis from x = 1 to x = 3 using lower rectangles of width 0.5 units. f(x) = 0.2x2 + 3 THINK 1 2

3

4 5

WRITE

3

Write the number of There are 4 rectangles of x 0 0.5 1 1.5 2 2.5 3 rectangles and their width. width 0.5 units. Find the height of each h1 = f(1) = 0.2(1)2 + 3 = 3.2 rectangle (left) by substituting h2 = f(1.5) = 0.2(1.5)2 + 3 = 3.45 the appropriate x-value into h3 = f(2) = 0.2(2)2 + 3 = 3.8 the f(x) equation. h4 = f(2.5) = 0.2(2.5)2 + 3 = 4.25 Area equals the width Area = width × (sum of heights of 4 rectangles) multiplied by the sum of the = 0.5(3.2 + 3.45 + 3.8 + 4.25) heights. = 0.5(14.7) Calculate this area. = 7.35 State the solution. The approximate area is 7.35 square units.

Chapter 9 Integration

363

The upper rectangle method Consider the area between the curve f(x) shown at right, y f(x) the x-axis and the lines x = 1 and x = 5. If the area is approximated by ‘upper rectangles’ of width 1 unit then the top of each rectangle is above the graph and touches the curve at one point. (In this case the top right-hand corner of the rectangle touches the graph.) R1 R 2 R 3 R 4 So, the height of R1 is f(2) units 0 1 2 3 4 5 x and the area of R1 is 1 × f(2) square units. Similarly, the area of R2 = 1 × f(3) square units, the area of R3 = 1 × f(4) square units the area of R4 = 1 × f(5) square units. Therefore, the approximate area between the curve f(x), the x-axis and the lines x = 1 to x = 5 is (R1 + R2 + R3 + R4) = 1[f(2) + f(3) + f(4) + f(5)] square units. If the same area was approximated with upper rectangle widths of 0.5 units, the sum of their areas would equal: 0.5[f(1.5) + f(2) + f(2.5) + f(3) + f(3.5) + f(4) + f(4.5) + f(5)] square units. From the diagram it can be seen that the upper rectangle approximation is greater than the actual area. Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation

WORKED Example 18 Find an approximation for the area in the diagram in worked example 17 using upper rectangles of width 0.5 units. f(x) = 0.2x2 + 3 THINK 1

WRITE

Find the number of rectangles and the height of each one (from left to right).

There are 4 rectangles: h1 = f(1.5) = 0.2(1.5)2 + 3 = 3.45 h2 = f(2) = 0.2(2)2 + 3 = 3.8 h3 = f(2.5) = 0.2(2.5)2 + 3 = 4.25 h4 = f(3) = 0.2(3)2 + 3 = 4.8

3

Area is the width of the interval multiplied by the sum of the heights. Calculate the area.

4

State the solution.

2

Area = 0.5(3.45 + 3.8 + 4.25 + 4.8) = 0.5(16.3) = 8.15 The approximate area is 8.15 square units.

Math

cad

Math

cad

It can be seen that the lower rectangle approximation (7.35 units) is less than the upper rectangle approximation (8.15 units). If the area is divided into narrower strips, the estimate of the area would be closer to the true value. Use one of the Mathcad files to investigate the effect of a greater number of strips in a given interval.

Right boxes

Left boxes

364

Mathematical Methods Units 3 and 4

The trapezoidal method

a

h Recall that the area of a trapezium = --- (a + b). 2 h The trapezoidal method involves a series of straight line approximations to the curve which generates strips in b the shape of trapeziums. Consider the area under the graph of f(x) between the x-axis and the lines x = 1 to x = 5. For each trapezium the width, or height, h = 1 unit. y For T1, a = f(1) and b = f(2) f(x) For T2, a = f(2) and b = f(3) For T3, a = f(3) and b = f(4) T1 T2 T3 T4 For T4, a = f(4) and b = f(5) The area of T1 = 1--2- [f(1) + f(2)]

0 1 2 3 4 5 x

+ f(3)] and so on. The area of T2 = The total area of the trapeziums is: 1 --- [f(1) + f(2) + f(2) + f(3) + f(3) + f(4) + f(4) + f(5)] 2 = 1--2- [f(1) + 2f(2) + 2f(3) + 2f(4) + f(5)] square units. The first and last terms are counted only once but all others are counted twice. 1 --- [f(2) 2

WORKED Example 19

Find an approximation for the area enclosed by the graph of f(x) = 0.2x2 + 3, the x-axis and the lines x = 1 to x = 3 using interval widths of 0.5 units and using the trapezoidal method. THINK 1 Sketch the graph of f(x). 2 Draw trapeziums of width 0.5 unit from x = 1 to x = 4.

WRITE/DRAW y 3

–2 3

Evaluate the height of each vertical side of the trapeziums by substituting the appropriate x-value into f(x).

4

Calculate the area by using the formula for the area of a trapezium where h is the width of the interval.

0 1 2 3 4 x

f(1) = 0.2(1)2 + 3 = 3.2 f(1.5) = 0.2(1.5)2 + 3 = 3.45 f(2) = 0.2(2)2 + 3 = 3.8 f(2.5) = 0.2(2.5)2 + 3 = 4.25 f(3) = 0.2(3)2 + 3 = 4.8 Total area of trapeziums =

0.5 ------- (3.2 2

+ 2 × 3.45 + 2 × 3.8 + 2 × 4.25 + 4.8)

= 0.25 × 31 = 7.75 Therefore, the area under the curve is approximately 7.75 units.

Note that the lower rectangle approximation found in worked example 17 was 7.35 units and the upper rectangle approximation found in worked example 18 was 8.15 7.35 + 8.15 units. The average of these two approximations is --------------------------- or 7.75 units which is 2 the same as the trapezoidal approximation for the area.

Chapter 9 Integration

365

WORKED Example 20 With width intervals of 1 unit calculate an approximation for the area between the graph of f(x) = x2 + 2 and the x-axis from x = −2 to x = 3 using: a lower rectangles b upper rectangles c averaging of the lower and upper rectangle areas. THINK 1 Sketch the graph of f(x) over a domain which exceeds the width of the required area. 2 Draw the lower and upper rectangles.

WRITE y

y = x2 + 2

2 –2 –1

0 1 2 3 x = Upper rectangles = Lower rectangles

a

1

2

b

Calculate the height of the lower a Lower rectangle heights: rectangles by substituting the f(−1) = (−1)2 + 2 = 3 appropriate values of x into the f(0) = 02 + 2 = 2 equation for f(x). Note the two f(0) = 2 rectangles to the right and left of f(1) = 12 + 2 = 3 the origin have the same height f(2) = 22 + 2 = 6 and are equal in area. Find the area by multiplying Area = 1(3 + 2 + 2 + 3 + 6) the width by the sum of the = 16 heights. Using lower rectangles, the approximate area is 16 square units.

1

Calculate the height of the upper rectangles by substituting the appropriate values of x into the equation for f(x).

2

Find the area by multiplying the width by the sum of the heights.

c Find the average by adding the area of the upper rectangles and lower rectangles and dividing by 2.

b Upper rectangle heights: f(−2) = (−2)2 + 2 = 6 f(−1) = 3 (from above) f(1) = 3 f(2) = 6 f(3) = 32 + 2 = 11 Area = 1(6 + 3 + 3 + 6 + 11) = 29 Using upper rectangles, the approximate area is 29 square units. 16 + 29 c Average of the areas = -----------------2 = 22.5 The approximate area is 22.5 square units when averaging the upper and lower rectangle areas and using widths of 1 unit.

Note that this average is between the area of the upper rectangles and the area of the lower rectangles and is closer to the actual area.

366

Mathematical Methods Units 3 and 4

remember remember 1. An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles or trapeziums which are all the same width. The approximation is found by finding the sum of all the areas of the rectangles or trapeziums. 2. Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation 3. Trapezoidal approximation = lower rectangle approximation + upper rectangle approximation --------------------------------------------------------------------------------------------------------------------------------------------------------2

Approximating areas enclosed by functions

9D WORKED

Example

17

1 Find an approximation for the area between the curve f(x) at right and the x-axis from x = 1 to x = 5 using a lower rectangle of width 2 units.

y

(3, 3) 3 (1, 2) 2

2 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. a b y y (5, 4)

19

(1, 2) 12 f(x) 0

1

5

x

5

3

x

b The height of the right-hand rectangle is: A 9 units B 4 units C 16 units D 12 units E 1 unit

(4, 19) (3, 12)

7 (2, 7) 4 (1, 4) 3 0 1 2 3 4 5x

3 multiple choice Consider the graph of y = x2 from x = 0 to x = 4 at right. a The width of each rectangle is: A 1 unit B 2 units C 3 units D 4 units E varying

c

0 1

f(x)

4 2

f(x)

y

y = x2

0

1 2 3 4

x

The area between the curve y = x2 and the x-axis from x = 0 to x = 4 can be approximated by the area of the lower rectangles as: A 20 sq. units B 14 sq. units C 18 sq. units D 15 sq. units E 30 sq. units

367

Chapter 9 Integration

WORKED

GC pro

x

1 2 3 4

5 Find an approximation for the area enclosed by the graph of f(x) = x2, the x-axis and the lines x = 1 and x = 3 with interval widths of 1 unit using the trapezoidal method.

y

Left and right box estimations of area Math

y = x2

cad

19

Left boxes

gram

Example

Math

Right boxes

0

WORKED

Math

y = x2

y

cad

18

4 a Find an approximation for the area in the diagram at right using upper rectangles of width 1 unit. b A better approximation for the area under this curve can be found by averaging the upper and lower rectangle areas. State this approximate value.

cad

Example

Trapezoidal rule

0

1

2

gram

GC pro

x

3

Trapezoidal rule

6 Find the approximate area between the curves below and the x-axis, over the interval indicated, by calculating the area of the shaded trapeziums. a b y y f(x) f(x) (4, 5)

5 3

(2, 3)

0

3 2

2

(6, 5)

5

x

4

(4, 3) (2, 2)

0

4

2

6

x

7 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. a y b y c y d y (2, 11) (1, 8)

8 6

x 5 f(x)

0 1

11 10

(1, 8) (3, 8)

8 7

(3, 10)

8 7

(4, 7)

x 5 f(x)

0 1 3

e

(4, 10)

10 9 6 4 0

(3, 9)

4

g f(x)

(1, 4)

f(x)

(2, 6)

3

5

x

y 7

(1, 7)

(4, 7) (2, 5)

5 4

(3, 2)

2

0 1

x

1 2 3 4 5

f y

y

f(x)

(3, 3)

3

f(x) 0

(5, 5)

5

(1, 4) 1

2 3 4 5

x

0

1

3

5

x 0

1

2

3

4

5

f(x) x

368 WORKED

Example

20

Mathematical Methods Units 3 and 4

8 With width intervals of 1 unit calculate an approximation for the area between the graph of f(x) = x2 + 4 and the x-axis from x = 1 to x = 4 using: a lower rectangles b upper rectangles c averaging of the lower and upper rectangle areas.

y y = x2 + 4

9 Find the approximate area between the curves below and the x-axis, over the interval indicated, by calculating the area of the shaded rectangles. Give exact answers. ay b c y y y = ex

y = log ex

y = –x2 + 3x + 8

0

1 2 3 4 x = 1 to x = 4

x –1 0 1 2 x = –1 to x = 2

x

e y

f f(x) =

1– 3 3x

0

g

y

y = (x – 4)2

0

y

x

0 1 2 3 4 5 x = 1 to x = 5

0

1 2 3 4 5 6 x = 2 to x = 6

x

x = 2 to x = 6 2 3 4 5 6 x

0

–3 –2.5 –2 –1.5 –1

x

2 3 4

dy

0 1 2 3 4 5 x x = 1 to x = 5

f(x) = –x2 – 4x

– 3x2 + 8x

1

y = x3 – 6x2

x

x = –3 to x = –1

10 Find an approximation for the area between the graphs below, the x-axis, and the lines x = 1 and x = 5, using interval widths as shown in the diagrams below. Give exact answers. a y b c y y 2 15 14 13 10

(3, 15)

(2, 14)

y = –x + 6x – 5

(4, 13)

(1, 10)

–1

f(x) 0

1 3 y=— 2x

1

d

–1

2 3 4 y y = 10 – x2

– 1–2

01 1 –

0

1

2

3

4

5

x

x

e

y

y = ex

x

2

0

1

2

3

x

0

1 2 3

x

Chapter 9 Integration

369

11 Calculate an approximation for the area between the graph of y = x(4 − x), the x-axis and the lines x = 1 and x = 4, using interval widths of 1 unit and: a lower rectangles b upper rectangles c averaging the lower and upper rectangle areas. 12 Calculate an approximation for the area under the graph of y = x2 − 4x + 5 to the x-axis between x = 0 and x = 3, using interval widths of 0.5 units and: a lower rectangles b upper rectangles c averaging the lower and upper rectangle areas. 13 Find an approximation for the area under the graph of y = 2x between x = 0 and x = 3, using interval widths of 1 unit, by averaging the lower and upper rectangle areas. 14 Find an approximation for the area between the graph of f ( x) = ( x − 1)3 and the x-axis, between x = 1 and x = 4, using the trapezoidal rule and: a interval widths of 1 unit b interval widths of 0.5 units. Give answers correct to 1 decimal place. 1 15 Calculate an approximation for the area under the graph of y = --- between x = 0.5 x and x = 2.5, using the trapezoidal rule and: a interval widths of 1 unit b interval widths of 0.5 units c interval widths of 0.25 units. Give answers correct to 2 decimal places. 16 Calculate an approximation for the area under the graph of y = 2 log e (x − 1) between x = 2 and x = 6 using the trapezoidal rule and interval widths of 1 unit.

r Park Rive 20 40 60 80

0

metres Block of land

100

17 At the back of a rectangular block of land, 100 metres long, is a park and a river. The distance to the river from the top of the rectangular block is shown in the table below.

Distance across rectangular block in metres

0

20 40 60 80 100

Distance of river from the block in metres

0

30 20 40 60

50

Calculate an approximation for the area of parkland between the rectangular block and the river by: a using the area of the ‘upper’ rectangles. b using the trapezoidal rule (use intervals of width 20 metres).

Mathematical Methods Units 3 and 4

18 Calculate an approximate area under the graph of f(x) = sin x, between x = 0 and π x = π, using the trapezoidal rule and interval widths of --- units. Give your answer 6 correct to 2 decimal places. 19 The graph below shows the velocity of a cyclist (in metres per second) at time t seconds after commencing a race. V Velocity (m/s)

370

(25, 14) (30, 15) (15, 12)

(20, 13)

(10, 10) (5, 5) 0

10

20

30 40 Time (s)

t

a What does the shaded area represent? b Find the approximate distance travelled by the cyclist in the first 30 seconds using the trapezoidal rule and interval widths of 5 seconds.

20 Answer the following statements concerning approximate areas under graphs as True or False. a An approximation for the area can be found quickly if very small interval widths are used. b The smaller the interval width used, the more accurate the approximation for the area. c The upper rectangle method is always more accurate than the lower rectangle method. d Averaging the upper rectangle area and the lower rectangle area is more accurate than using the ‘upper’ or ‘lower’ approximations on their own.

Chapter 9 Integration

371

The fundamental theorem of integral calculus Consider the region under the curve f(x) between x = a and x = b, where f(x) ≥ 0 and is continuous for all x ∈ [a, b]. Let F(x) be the function that is the measure of the area under the curve between a and x. F(x + h) is the area under the curve between a and x + h and F(x + h) − F(x) is the area of the strip indicated on the graph. y

0

y = f(x)

a x x+h b F(x) F(x + h) – F(x)

x

The area of the strip is between the areas of the upper and lower rectangles; that is, f ( x )h < F ( x + h ) – F ( x ) < f ( x + h )h F( x + h) – F( x) or f ( x ) < -------------------------------------- < f ( x + h ) , h ≠ 0 (dividing by h). h As h → 0, f(x + h) → f(x) or

F ( x + h) – F( x) lim ---------------------------------------- = f ( x ) h

h→0

that is, F′(x) = f(x) (differentiation from first principles). Therefore,

∫

F(x) = f ( x ) dx

that is, F(x) is an antiderivative of f(x) or

∫ f ( x ) dx = F(x) + c

but when x = a,

∫ f ( x ) dx = F(a) + c = 0 (as the area defined is zero at x = a) or Therefore, and when x = b,

c = −F(a).

∫ f ( x ) dx = F(x) − F(a) ∫ f ( x ) dx = F(b) − F(a).

That is, the area under the graph of f(x) between x = a and x = b is F(b) − F(a).

372

Mathematical Methods Units 3 and 4

This is the fundamental theorem of integral calculus and it enables areas under graphs to be calculated exactly. It can be stated as: Area =

b

∫a f ( x ) dx

= [ F ( x ) ] ab [do not add c as F(x) is an antiderivative of f(x)] = F(b) − F(a) a and b are called the terminals of this definite integral.

∫a f ( x ) dx b

is called the definite integral because it can be expressed in terms of its

terminals a and b, which are usually real numbers. In this case the definite integral evaluates as a real number and not a function. The function being integrated, f(x), is called the integrand.

Definite integrals 1. Evaluate

1

∫1 2x dx 4

∫1 2x dx

2. Evaluate a

b

4

2

∫1 2x dx + ∫2 2x dx

c Compare answers a and b. 3. a 2

4

∫ 1 x dx

b Compare the answer to the answer to 2a. 3

∫0

4. Evaluate a

[ 2x + x 2 ] dx

b

3

∫0

2x dx +

c Compare answers a and b. 3

∫1

5. Evaluate a

b

2x dx

1

∫3 2x dx

c Compare answers a and b.

Properties of definite integrals Definite integrals have the following five properties. 1. 2. 3.

a

∫ a f ( x ) dx = 0 b

∫a b

∫a

b

f ( x ) dx =

c

∫a

kf ( x ) dx = k

f ( x ) dx +

∫ c f ( x ) dx , a < c < b

b

∫a f ( x ) dx b

4.

∫a

5.

∫a f ( x ) dx = –∫b f ( x ) dx

b

b

[ f ( x ) + g ( x ) ] dx = a

∫a

f ( x ) dx +

b

∫a g ( x ) dx

3

∫0 x 2 dx

Chapter 9 Integration

WORKED Example 21 Evaluate the following definite integrals. a b

3

∫0 ( 3 x 2 + 4 x – 1 ) d x 2

4

- dx ∫–1 (---------------------2 x + 1 )3

THINK

WRITE

a

a

1

Antidifferentiate each term of the integrand and write in the form [ F ( x ) ] ab .

3

∫0 ( 3x 2 + 4x – 1 ) dx = [ x 3 + 2x 2 – x ] 03

2

Substitute values of a and b into F(b) − F(a).

= [ 33 + 2( 3 )2 – 3 ] – [ 03 + 2( 0 )2 – 0 ]

3

Evaluate the integral.

= 42 − 0 = 42

b

1

Express the integrand in simplest index form.

2

Antidifferentiate by rule.

b

2

4

2

dx = ∫ 4 ( 2x + 1 ) –3 dx ∫–1 (---------------------2x + 1 ) 3 –1 2

[ 4 ( 2x + 1 ) –2 ] –1 = -------------------------------------2 × –2 2

= [ – ( 2x + 1 ) –2 ] –1 2

Express the integral with a positive index number.

–1 = ----------------------2 ( 2x + 1 )

4

Substitute values of a and b into F(b) − F(a).

–1 1 = – ----2- – ------------2( –1 ) 5

5

Evaluate the integral.

1 = – ------ + 1 25

3

24 = -----25

–1

373

374

Mathematical Methods Units 3 and 4

WORKED Example 22 Find the exact value of each of the following definite integrals. a

2π

x

∫π sin --6- d x

b

1

∫– 1 ( e 3 x – e – 3 x ) d x

THINK

WRITE

a

a

1

Antidifferentiate the integrand, writing it in the form [ F ( x ) ] ab .

2

Substitute values of a and b into F(b) − F(a).

3

2π

∫π

x x sin --- dx = – 6 cos --6 6

2π

π

π 2π = – 6 cos ------ – – 6 cos --6 6 π π = – 6 cos --- – – 6 cos --3 6 1 3 = – 6  --- – – 6  -------  2  2

Evaluate the integral.

= [ –3 ] – [ –3 3 ] = –3 + 3 3 b

1

Antidifferentiate the integrand, using [ F ( x ) ] ab .

1

b

∫–1 ( e3 x – e–3 x ) dx 1

= [ 1--3- e 3 x + 1--3- e –3 x ] –1

2

Substitute values of a and b into F(a) − F(b).

= [ 1--3- e 3 + 1--3- e –3 ] – [ 1--3- e –3 + 1--3- e 3 ]

3

Evaluate.

= 1--3- e 3 + 1--3- e –3 – 1--3- e –3 – 1--3- e 3 =0

Graphics Calculator tip! Finding definite integrals It is possible to find the numerical value for the definite integral using the graphics calculator. Using the TI–83, follow these steps to evaluate 1. 2. 3. 4. 5.

2π

∫ π sin  --6- dx. x

Ensure radian MODE is selected. Enter the function sin(X/6) as Y1 in the Y= menu. Press 2nd [QUIT]. Press MATH and select 9:fnInt( Complete the expression as shown at right. (Y1 is found under VARS /Y-VARS/1:Function.

The answer is 2.196 15 which is the answer correct to 5 decimal places. This is the same as −3 + 3 3 found in worked example 22.

Chapter 9 Integration

375

WORKED Example 23 If

k

∫0 8 x d x = 36 , find k.

THINK

WRITE k

1

∫0 8x dx = [ 4x 2 ]0

Antidifferentiate the integrand, using [ F ( x ) ] ab .

k

k

So [ 4x 2 ] 0 = 36 2

Substitute the values of a and b into F(a) − F(b).

3

Simplify the integral.

[4k2] − [4(0)2] = 36 4k2 − 0 = 36 4k2 = 36 k2 = 9 k=±

4

9

k = 3 or −3

Solve the equation.

remember remember 1. The fundamental theorem of calculus is

b

∫a f ( x ) dx = [ F ( x ) ]a

= F(b) − F(a).

where F(x) is an antiderivative of f(x). 2. The expression

b

b

∫a f ( x ) dx is called the definite integral where a and b are the

terminals and represent the upper and lower values of x. 3. Properties of definite integrals (a)

a

∫ af ( x ) dx = 0 b

c

∫a

(c)

∫a kf ( x ) dx = k ∫a f ( x ) dx

(d)

∫ a [ f ( x ) + g ( x ) ] dx = ∫ a f ( x ) dx + ∫ a g ( x ) dx

(e)

∫ a f ( x ) dx = –∫ b f ( x ) dx

f ( x ) dx =

∫a

b

(b)

f ( x ) dx +

b

b

b

b

∫ c f ( x ) dx , a < c < b

b

a

b

376

Mathematical Methods Units 3 and 4

9E 9.2

SkillS

HEET

WORKED

Example

21

d hca

1 Evaluate the following definite integrals. a

Mat

d Integrator

g

1

∫0 ∫

x 2 dx

b

1 ----2- dx x 2

e

6

1

∫– 1 2

WORKED

Example

22

The fundamental theorem of integral calculus 3

∫0 2

∫0

x 3 dx

c

( x 3 + 3x 2 – 2x ) dx

f

–2

∫ –4

(6 – 2x + x 2 ) dx

h

( 4x –2 + 2x – 6 ) dx

k

∫0

n

- dx ∫0 --------------------( 2x – 7 ) 3

j

∫1

m

∫–2 –4 ( 2 – 3x )3 dx

p

- dx ∫ 1 ---------------------x

s

- dx ∫3 -----------------2x – 5

0

3 2x 3

7

+ 5x 2

1

3

( x 3 + x – 4 ) dx

i

4

∫ 3 (x 2 – 2x) dx 4

∫1

1 ---

(3x 2 + 2x 2 ) dx

9

∫ 43

x dx

2

l

∫–1 3 ( 5x – 2 )4 dx

5

o

∫ 1 ( 2x

q

- dx ∫1 ----5x

r

- dx ∫ 0 --------------------( 3x – 4 ) 5

t

- dx ∫–2 -----------------8 – 3x

2 ( x + 4 ) 4 dx

3

5

3

0

4

3 --2

2

–4

– 3x –1 ) dx

6

2 Find the exact value of each of the following definite integrals. a d

g

j

m p

3 If

2

∫0

2

∫1 π --2

∫0

e 4 x dx

b

( 3e 6 x + 5x ) dx

e

sin x dx

h

2π

∫π

2π

∫π ∫ 5

π

x – 2 sin --- dx 3

k

x 3 cos --- dx 6

n

x π  2 + sin --- dx 4 0

q

0

∫– 2 ∫ ∫

x ---

e 3 dx

4

x

 5--- + e --2- dx  1x π π --2

3 sin 4x dx

0

∫– π ∫

c

i

∫0 5 sin --4- dx

x – 7 cos --- dx 2 –2 π

o

0

∫1

5

∫1 f ( x ) dx = 6 , find the value of ∫ 1 3f ( x ) dx .

–1

∫ –3 ( e2 x – e–2 x ) dx

l

( x 2 + 3 – 6 sin 3x ) dx

∫–1 –4e–2 x dx

f

cos 2x dx

3

1

r

π

∫

π --2 –π -----2

x

8 cos 4x dx

π – --2

∫ –π –4 cos 3x dx π

∫ 1  --x- + 3 cos --2- dx 1

x

Chapter 9 Integration

4 multiple choice 5

Given that a

b

∫ 1 f ( x ) dx = 6 ,

5

∫ 1 [ f ( x ) + 1 ] dx is equal to: A 16

B 10

D 19

E 22

C 11

1

∫5 f ( x ) dx is equal to: A −6

B 5

D 6

E 0

C −5

5 Evaluate the following. a

c

WORKED

Example

3

∫0

( t 2 – 4t ) dt

7

3

- dp ∫4 -----------------( p – 3) π

x ---

∫0

g

- dt ∫ 1 ------t t

23

7 If

8 If

9 If

10 If

d

3 --2

e

6 If

b

( e 4 – cos 2x ) dx

4 –3

π --2

∫ 0 2 cos 3t dt 10 π

∫5 π  –sin --5- dx 2

x

8

f

- dm ∫ 1 --------------4m – 3

h

∫–1 ( 3 sin 2x – e–3 x ) dx

1

k

∫ 0 ( 2x + 3 ) dx = 4 , find k. k

∫0 3x 2 dx = 8 , find k. k2

∫ 1 --x- dx = loge 9 , find k. a

∫0

x ---

e 2 dx = 4 , find the value of a.

π

3

π

∫ k cos 2x dx = – ------4- , find the value of k given that 0 < k < --2- .

377

378

Mathematical Methods Units 3 and 4

Signed areas When calculating areas between the graph of a function f(x) and the x-axis using the definite integral

b

∫a f ( x ) dx , the area is signed; that is, it is positive or negative. If f(x) > 0,

the region is above the x-axis; if f(x) < 0 it is below the axis. We shall now examine these two situations and look at how we calculate the area of regions that include both.

Region above axis

y

y = f(x)

If f(x) > 0, that is, the region is above the x-axis, then b

∫ a f ( x ) dx > 0 , so the value of the definite integral is positive. For example, if f(x) > 0, then the area =

b

∫a f ( x ) dx .

0

a

x

b

Region below axis If f(x) < 0, that is, the region is below the x-axis, then

y

y = f(x)

b

∫a f ( x ) dx < 0 , so the value of the definite integral is negative.

a

b x

0

For example, if f(x) < 0, then the area = − region is below the x-axis or area =

b

∫a f ( x ) dx , as the

a

∫ b f ( x ) dx

(reversing the

terminals changes the sign). Therefore, for areas below the x-axis, the sign of the definite integral must be made negative, or the terminals reversed, to ensure that the area has a positive value. (Areas cannot be negative.)

Combining regions

y

For regions which are combinations of areas above and below the x-axis, each area has to be calculated by separate integrals — one for each area above and one for each area below the x-axis. For example, from the diagram, Area = A1 + A2 However,

A1 a

0 A 2

b

∫a f ( x ) dx = A1 – A2 , because the areas are signed.

To overcome this difficulty we find the correct area as: Area = or

=

b

∫c

b

f ( x ) dx –

c

∫ a f ( x ) dx c

∫ c f ( x ) dx + ∫ a f ( x ) dx

y = f(x)

(= A1 − −A2 = A1 + A2)

c

b

x

Chapter 9 Integration

379

Note: When calculating the area between a curve and the x-axis it is essential that the x-intercepts are determined and a graph of the curve is sketched over the interval required. The term |x| means make the value of x positive even if it is negative.

WORKED Example 24

a Express the shaded area as a definite integral. b Evaluate the definite integral to find the shaded area, giving your answer as an exact value.

y

0

THINK

WRITE

a Express the area in definite integral notation where a = 1 and b = 4.

a Area =

b

b Area = [ log e x ]

1

Antidifferentiate the integrand.

2

Evaluate.

3

State the solution as an exact answer.

y = –x1

1

4

x

41

∫1 --x- dx 4 1

= [log e 4] – [log e 1] = log e 4 − 0 = log e 4 The area is log e 4.

WORKED Example 25

Calculate the shaded area.

y y = x2 – 4x 0

1

3 4 x

THINK 1

2 3

Express the area in definite integral notation, with a negative sign in front of the integral as the region is below the x-axis. Antidifferentiate the integrand. Evaluate.

WRITE Area = –

3

∫1( x

2

– 4x ) dx 3

= – [ 1--3- x 3 – 2x 2 ] 1

= – { [ 1--3- ( 3 ) 3 – 2 ( 3 ) 2 ] – [ 1--3- ( 1 ) 3 – 2 ( 1 ) 2 ] } = – { [ 9 – 18 ] – [ 1--3- – 2 ] } = – [ –9 – ( – 1 2--3- ) ] = – [ –9 + 1 2--3- ] = – ( – 7 1--3- ) = 7 1--3-

4

State the solution.

The area is 7 1--3- square units.

380

Mathematical Methods Units 3 and 4

WORKED Example 26

a Express the shaded area as a definite integral. b Calculate the area.

y y = x3 – 4x 0

–2

2

THINK

WRITE

a Express the area above the x-axis as an integral and the area below the x-axis as an integral. For the area below the x-axis, take the negative of the integral from 0 to 2.

a Area =

b

b = [ 1--4- x 4 – 2x 2 ] –02 – [ 1--4- x 4 – 2x 2 ] 02

1

Antidifferentiate the integrands.

2

Evaluate.

0

∫– 2

( x 3 – 4x ) dx –

x

2

∫ 0 ( x3 – 4x ) dx

= {[ 1--4- (0)4 − 2(0)2] − [ 1--4- (−2)4 − 2(−2)2]} − {[ 1--4- (2)4 − 2(2)2] − [ 1--4- (0)4 − 2(0)2]}

3

Simplify.

4

State the solution.

= [0 − (4 − 8)] − [(4 − 8) − 0] = 4 − (−4) =8 The area is 8 square units.

0

To evaluate

∫– 2

( x 3 – 4x ) dx –

2

∫0 ( x 3 – 4x ) dx using

a graphics calculator, use the fnInt command found under MATH 9:fnInt( for each part.

WORKED Example 27

a Sketch the graph of y = ex − 2 showing all intercepts and using exact values for all key features. b Find the area between the curve and the x-axis from x = 0 to x = 2. THINK a

1

Find the x-intercept by letting y = 0 and solving for x. Take log e of both sides.

WRITE a When y = 0, ex – 2 = 0 ex = 2 log e e x = log e 2 x = log e 2 (or approximately 0.69) so the x-intercept is log e 2.

381

Chapter 9 Integration

THINK

WRITE y = e0 − 2 =1−2 = −1 so the y-intercept is −1.

When x = 0,

2

Find the y-intercept by letting x = 0.

3

Note the vertical translation and hence sketch the graph showing the appropriate horizontal asymptote and intercept. Shade the region required.

4

y y = ex – 2

–2

0 –1

loge2

–2

b

1

Express the area above the x-axis as an b Area = integral and the area below the x-axis as an integral. Subtract the area below the x-axis from the area above the x-axis.

2

∫log 2

x 2 y=–2

( e x – 2 ) dx −

e

loge 2

∫0

( e x – 2 ) dx

log e 2

2

Antidifferentiate the integrands.

2 = [ e x – 2x ] log – [ e x – 2x ] 0 2

3

Evaluate. (Remember: elog e a = a)

4

Simplify.

= [e2 − 2(2)] − [elog 2 − 2 log e 2] − {[eloge 2 − 2(log e 2)] − [e0 − 2(0)]} = [e2 − 4] − [2 − 2 log e 2] − {[2 − 2 log e 2] − [1 − 0]} = e2 − 4 − 2 + 2 log e 2 − 2 + 2 log e 2 + 1 = e2 − 7 + 4 log e 2

5

State the solution.

e

The area is (e2 − 7 + 4 log e 2) or approximately 3.162 square units.

remember remember 1. If f(x) > 0, area =

b

∫a f ( x ) dx .

2. If f(x) < 0, area = − =

b

∫a f ( x ) dx , as the region is below the x-axis or a

∫b f ( x ) dx , (reversing the terminals changes the sign).

3. If the required area lies above and below the x-axis: (a) find the intercepts and sketch the graph (b) calculate the integrals and subtract the area below the x-axis from the area above the x-axis. 4. Area = or

=

b

y

y = f(x)

c

∫c f ( x ) dx – ∫a f ( x ) dx (= A1 − −A2 = A1 + A2) b

∫c f ( x ) dx + ∫a f ( x ) dx

5. e loge a = a

A1

c

a

0 A 2

c

b

x

382

Mathematical Methods Units 3 and 4

9F

Signed areas y

1 Find the area of the triangle at right. a geometrically b using integration

y=x

0 y

2 Find the area of the triangle at right. a geometrically b using integration

3

0 WORKED

Example

24a

x

4

x

3 y=3–x

3 Express the following shaded areas as definite integrals. a

y

b y

y = 2x

c

y

01

e

y

3

x 4 y=4–x

0

x

y = x3 – 9x2 + 20x

f

0

1 2

y

g

y

d

y = x2

4

y

x

hy

1 0

–2 0

i

y 2

1

3

x

WORKED

Example

24b

π – 2

x

–1

0

x

1

0

1

4

x

y y = cos –3x

1 0

y = e–2x

x

y = –x3 – 4x2 – 4x

j

y = 2 sin 2x

x

–1 0

–3

y = ex

y = 3x2

0

x

3— π 2

4 Evaluate each of the definite integrals in question 3 to find the shaded area. Give your answer as an exact value. 5 For each of the following, sketch a graph to illustrate the region for which the definite integral gives the area.

Mat

d hca

Definite integrals

a

3

∫ 0 4x dx 1

d

∫– 1

g

4

∫

( 4 – x 2 ) dx

x log e --- dx 2 2

b e h

2

∫ 1 ( 6 – x ) dx 4

∫1

π --2

x dx

∫ 0 3 sin 2x dx

c f

3

∫ 1 x 2 dx 0

∫–3 2e x dx

383

Chapter 9 Integration

WORKED

Example

25

6 Calculate each of the shaded areas below. y b c a y

0

d

x

–1 0

–2

x

2

y y = x2 – 4

y = –4 – 2x

y=x–2

0

2

y

–2

x

x

0

–1

–2 y = 1– x2

e

y

f

y = x3

g

y

0

–1 x

0

–2

h

y

1 x

0

y = –e–2x

y = –ex

y = x3 + 2x2 – x –2

j

1– 2

x

0 1 x

–1

iy

1

y

y

y = sin x 0

π

3π — 2

x

–π 0

–2π

x

2π

y = 2 cos –2x

7 multiple choice y

a The area between the graph, the x-axis and the lines x = −2 and x = −1 is equal to: 2

∫1

A

–1

f ( x ) dx

B

1

∫2 f ( x ) dx

D

∫ –2

E −

0

f ( x ) dx

C

∫–1 f ( x ) dx

0

–2

3 y = f(x)

–1

∫ –2 f ( x ) dx

x

b The area between the graph, the x-axis and the lines x = −2 and x = 3 is equal to: 3

A

∫0

D

∫3 f ( x ) dx

f ( x ) dx +

–2

∫0

3

∫– 2

f ( x ) dx B

–2

WORKED

Example

26a

1

f ( x ) dx

C

0

∫– 2

f ( x ) dx +

3

∫1 f ( x ) dx

3

∫–2 f ( x ) dx – ∫0 f ( x ) dx

E

8 Express the following shaded areas as definite integrals which give the correct area. y y y b c a g(x) f(x) 0

d

–5 –4 –2

2

5

x –3

e

y

0

f(x)

0 1

3

y

x –3 –2

x

–3 g(x)

0

2 3

x

–1

0 2

x h(x)

384

Mathematical Methods Units 3 and 4

9 multiple choice Examine the graph.

y

–2

0

1

3

x

y = x3 – 2x2 – 5x + 6

a The area between the curve and the x-axis from x = −2 and x = 1 is equal to: 1 - sq. units A 17 ----12

B 15 3--4- sq. units

D −15 3--4- sq. units

E 10 sq. units

1 - sq. units C −17 ----12

b The area between the curve and the x-axis from x = 1 and x = 3 is equal to:

c

WORKED

Example

27

A −6 2--3- sq. units

B 2 sq. units

D 5 1--3- sq. units

E 6 2--3- sq. units

C −5 1--3- sq. units

The area between the curve and the x-axis from x = −2 and x = 3 is equal to: 5 - sq. units A 10 ----12

B 11 3--4- sq. units

D 12 sq. units

1 - sq. units E 21 ----12

5 - sq. units C 22 ----12

10 Sketch the graph of the curve y = x2 − 4, showing all intercepts and using exact values for all key features. Find the area between the curve and the x-axis: a from x = 0 to x = 2 b from x = 2 to x = 4 c from x = 0 to x = 4. 11 Sketch the graph of the curve y = x3 + x2 − 2x, showing all intercepts. Find the area between the curve and the x-axis between the lines: a x = −2 and x = 0 b x = 0 and x = 1 c x = −2 and x = 1. 12 Sketch the graph of the curve y = 1 + 3 cos 2x over [0, π]. Find the exact area between the curve and the x-axis from: π a x = 0 to x = --4 3π b x = ------ to x = π. 4

Work

ET SHE

9.2

13 Sketch the graph of f(x) = x – 1 and find the area between the curve and the x-axis and the lines x = 2 and x = 3. Give both an exact answer and an approximation to 3 decimal places. 1 14 Find the exact area between the curve y = --x- , the x-axis and the lines x = 1--2- and x = 2. 15 Find the exact area bounded by the curve g(x) = e x + 2, the x-axis and the lines x = −1 and x = 3.

385

Chapter 9 Integration

Further areas

y y = f(x)

Areas bound by a curve and the x-axis For graphs with two or more x-intercepts, there is an enclosed region (or regions) between the graph and the x-axis. The area bound by the graph of f(x) and the x-axis is: –

0

a

x

b

b

∫a f ( x ) dx (negative because the area is below the x-axis).

y

The area bound by the graph of g(x) and the x-axis is: b

c

∫c g ( x ) dx – ∫a g ( x ) dx

a

c

0

b

x

That is, if the graph has two x-intercepts then one integrand y = g(x) is required. If the graph has three x-intercepts then two integrands are required, and so on. Note: Wherever possible it is good practice to use sketch graphs to assist in any problems involving the calculation of areas under curves.

WORKED Example 28

a Sketch the graph of the function g(x) = (3 − x)(2 + x). b Find the area bound by the x-axis and the graph of the function. THINK a 1 Determine the type of graph by looking at the number of brackets and the sign of the x-terms. 2 Solve g(x) = 0 to find the x-intercepts.

b

3

Sketch the graph.

1

Shade the region bound by g(x) and the x-axis.

WRITE a g(x) = (3 − x)(2 + x) is an inverted parabola. For x-intercepts, g(x) = 0 (3 − x)(2 + x) = 0 x = 3 and x = −2. The x-intercepts are −2 and 3. y

b x 3 y = g(x)

–2 0

2

Express the area as an integral.

3

Evaluate.

Area =

3

∫–2 ( 6 + x – x 2 ) dx

3 = [ 6x + 1--2- x 2 – 1--3- x 3 ] –2

= [6(3) + 1--2- (3)2 − 1--3- (3)3] = (18 +

− [6(−2) + 1--2- (−2)2 − 1--3- (−2)3]

9 --2

− 9) − (−12 + 2 + 8--3- )

= 13 1--2- − (−7 1--3- ) = 13 1--2- + 7 1--34

State the solution.

= 20 5--6The area bound by g(x) and the x-axis is 20 5--6square units.

386

Mathematical Methods Units 3 and 4

the area bound by a graph Graphics Calculator tip! Finding and the x-axis

▼

▼

The area bound by the graph and the x-axis can be drawn and evaluated using the graphics calculator. 1. Enter the function as Y1 in the Y= menu. 2. Press WINDOW and set Xmin to −4.7 and Xmax to 4.7. These values ensure the cursor reaches exact integers when tracing. 3. Press 2nd [CALC] 7: ∫ f(x)dx. 4. Press to the left x-intercept (−2), press ENTER , and press to the right x-intercept (3) and press ENTER . Note: This method can be used even if the area required is between non-intercepts.

Finding areas without sketch graphs When finding areas under curves which involve functions whose graphs are not easily sketched, the area can be calculated providing the x-intercepts can be determined. Note: The symbol |f(x)| is known as the absolute value of f(x) which means that we must make the f(x) function or value positive whenever it is negative. For example, |−5| = 5 |3| = 3 |− 2--3- | =

2 --3

As areas cannot be negative, taking the absolute value of the integrands involved in a problem will ensure that all areas are made positive. For example, b

or

=

y

c

∫ c f ( x ) d x + ∫a f ( x ) d x

The shaded area at right =

b

∫c

f( x) d x +

c

∫a

f( x) d x

y = f(x) a

c 0

b

WORKED Example 29

a Find the x-intercepts of y = sin 2x over the domain [0, 2π]. b Calculate the area between the curve, the x-axis and x = 0 and x = π. THINK a 1 To find the x-intercepts, let y = 0. 2 Solve for x over the given domain.

b

1

2

WRITE a For x-intercepts, y = 0 sin 2x = 0 2x = 0, π, 2π, 3π, 4π, etc. π 3π x = 0, ---, π, ------, 2π 2 2 π Pick the x-intercepts that are between b x = --- is the only x-intercept between 0 and π. 2 the given end points of the area.

State the regions for which it is necessary to calculate the area.

Area =

π --2

π

∫0 sin 2x dx + ∫--π2- sin 2x dx

x

Chapter 9 Integration

THINK 3

387

WRITE π --2

π

= [ – 1--2- cos 2x ] 0 + [ – 1--2- cos 2x ] π---

Evaluate the absolute value of the integral for each region.

2

=

[ – 1--2-

cos π –

( – 1--2-

cos 0 ) ]

+ [ – 1--2- cos 2 π – ( – 1--2- cos π ) ] = [ 1--2- – ( – 1--2- ) ] + [ – 1--2- – ( 1--2- ) ] 4

Add the result to give the total area.

5

State the solution.

= 1 + –1 =1+1 =2 The area is 2 square units.

WORKED Example 30

a Differentiate log e (x2 − 1).

x -. b Hence, find an antiderivative of ------------2 x –1 c

x - , the x-axis, x = 2 and x = 3, giving your Find the area between the graph of ------------x2 – 1 answer correct to 2 decimal places.

THINK

WRITE

a

a

1

Let y equal the expression to be differentiated.

2

Express u as a function of x in order to apply the chain rule for differentiation. (Let u equal the function inside the brackets.)

Let y = log e (x2 − 1) Let u = x2 − 1

du Find ------ . dx Write y in terms of u.

du ------ = 2x dx y = log e u

5

dy Find ------ . du

dy 1 ------ = --du u

6

dy Find ------ using the chain rule. dx

3 4

dy 1 So ------ = --- × 2x dx u 2x = ------------x2 – 1 Continued over page

388

Mathematical Methods Units 3 and 4

THINK b

1

2

3

WRITE

dy Since ------ dx = y + c , express the dx relationship in integral notation.

∫

dy Remove a factor from ------ so that it dx resembles the integral required. Divide both sides by the factor in order to obtain the required integral.

b

2x

- dx = log e ( x 2 – 1 ) + c ∫ ------------x2 – 1 x - dx = log e ( x 2 – 1 ) + c 2 ------------2 x –1

∫

x

- dx = ∫ ------------x2 – 1

1 --2

log ( x 2 – 1 ) + c

x - is An antiderivative of ------------2 x –1 c

1

Find the x-intercepts. x - = 0, the numerator = 0.) (For ------------2 x –1

2

If the x-intercepts are not between the terminals of the area, find the area by evaluating the integrand.

1 --2

log (x2 − 1).

x - =0 c For x-intercepts, ------------2 x –1 x=0 3

x

- dx ∫2 ------------x2 – 1

Area =

3

= [ 1--2- log e ( x 2 – 1 ) ] 2

= [ 1--2- log e ( 3 2 – 1 ) ] – [ 1--2- log e ( 2 2 – 1 ) ] =

1 --2

log e 8 – 1--2- log e 3

=

1 --2

log e 8--3-

= 3

State the solution.

1 --2

log e 8--3-

The area is

1 --2

log e 8--3- or approximately 0.49

square units.

remember remember 1. For graphs with two or more intercepts, there is an enclosed region (or regions) between the graph and the x-axis. 2. The number of regions is one less than the number of intercepts. 3. Where possible, sketch graphs to make it easier to calculate the areas under curves, or use a graphics calculator. 4. As areas can’t be negative, take the absolute values of the integrals. 5. When graphs are not easily drawn, areas can be calculated by finding the x-intercepts and determining whether they are within the bounds of the required area.

389

Chapter 9 Integration

9G

Further areas

In the following exercise give all answers correct to 2 decimal places where appropriate, unless otherwise stated. WORKED

Example

28

1

i Sketch the graph of each of the following functions. ii Find the area bound by the x-axis and the graph of each function. Use a graphics calculator to assist. a f(x) = x2 − 3x

b g(x) = (2 − x)(4 + x)

2 Find the area bound by the x-axis and the graph of each of the following functions. b h(x) = x2 + 5x − 6

c

Math

cad

a h(x) = (x + 3)(5 − x)

d g(x) = x3 − 4x2

g(x) = 8 − x2

Definite integrals

e f(x) = x(x − 2)(x − 3)

f

g g(x) = x + 3x − x − 3

h h(x) = (x − 1)(x + 2)(x + 5)

3

2

f(x) = x − 4x − 4x + 16 3

2

3 multiple choice The area bound by the curve with equation y = x2 − 6x + 8 and the x-axis is equal to: A 1 1--3- sq. units

B 6 2--3- sq. units

D 3 sq. units

E −1 1--3- sq. units

C 12 sq. units

4 multiple choice The area between the curve at right, the x-axis and the lines x = −3 and x = 4 is equal to: A

4

∫–3 f ( x ) dx

B

2

4

∫–3 f ( x ) dx + ∫2 f ( x ) dx

y

–3

0

2

4

x y = f(x)

C E

–3

f ( x ) dx

2

f ( x ) dx –

∫4

∫– 3

D

4

2

∫2 f ( x ) dx – ∫–3 f ( x ) dx

4

∫2 f ( x ) dx

5 multiple choice The area between the curve y = x2 − x − 6, the x-axis and the lines x = 2 and x = 4 is equal to: A 2 5--6- sq. units

B

2 --3

D 2 3--4- sq. units

E 4 1--2- sq. units

sq. units

C 5 sq. units

390

Mathematical Methods Units 3 and 4

6 For each of the following: i sketch the graph of the curve over an appropriate domain, clearly labelling any x-intercepts in the interval required (use a graphics calculator to assist). ii find the area between the curve, the x-axis and the lines indicated below. a y = 3 − 3x2, x = 0 and x = 2 c

Example

d y = x3 − 4x, x = −2 and x = 1 y = e−x, x = 0 and x = 2

e y = e2x, x = −2 and x = 0

f

π π g y = 2 sin x, x = --- and x = --6 3

x π h y = cos --- , x = --- and x = 2π 2 2

i WORKED

1 y = – ----2- , x = 1 and x = 2 x

2 b y = --- , x = 1 and x = 3 x

π π y = sin 3x, x = – --- and x = – --2 6

j

y = x x , x ≥ 0, x = 0 and x = 4

7 For each of the following functions: i find the x-intercepts over the given domain

29

ii calculate the area between the curve, the x-axis and the given lines. Use sketch graphs to assist your workings. a y = x − 4x−1, x ≠ 0, x = 1 and x = 3 b y = sin x − cos x, x ∈ [0, π], x = 0 and x = π c

y = e x − e, x = 0 and x = 3

1 d y = x − ----2- , x ≠ 0, x = x

1 --2

and x = 2

x ---

e y = e 2 , x = −2 and x = 2 f

y = x4 − 3x2 − 4, x = 1 and x = 4

g y = (x − 2)4, x = 1 and x = 3 8 Find the exact area of the region enclosed by the x-axis, y = e3x and the lines x = 1 and x = 2.

π 9 Find the exact area of the region enclosed by the x-axis, y = −cos x and the lines x = --3 5 π- . (Use a sketch graph to assist your calculation.) and x = ----6 10 Find the area bound by y = (x − 1)3, the x-axis and the y-axis. 1 11 a Sketch the graph of y = ------------------2- showing all asymptotes and intercepts. ( x – 3) b Find the area under the curve between x = −1 and x = 1. 12 a Give the equation of the asymptotes for the function f(x) = (x + 2)−3. b Find the area between the curve, the x-axis, x = −1 and x = 1.

Chapter 9 Integration

391

13 Find the area bound by the curve y = 3 − e2x, the x-axis, x = −2 and x = 0. (Find the x-intercepts first.)

π 14 Find the area bound by the curve y = 4 − sin 2x, the x-axis, x = – --- and x = π. 2 (Check the x-intercepts first.) WORKED

Example

30

15 a Differentiate x log e x. (x > 0) b Hence, find an antiderivative of log e x. c

Find the area bound by the graph of log e x, the x-axis, x = 1 and x = 4 giving exact answers.

16 a Differentiate log e (x2 + 2). x -. b Hence, find an antiderivative of ------------x2 + 2 c

x - , the x-axis, x = −1 and x = 1. Find the area between ------------x2 + 2

17 a Find the area between the graph of y = x2, the x-axis, x = 0 and x = 2. b Use this result to calculate the area between the graph, the y-axis and the line y = 4. y

y = x2 (2, 4)

0

x

2

18 Find the exact area of the shaded region on the graph y = e2x below. y y = e2x

0 2

x

19 Find the shaded area below. y

– π–2

0

π – 2

y = 2 sin x

x

392

Mathematical Methods Units 3 and 4

Areas between two curves We shall now consider the area between two functions, f(x) and g(x), over an interval [a, b]. Our approach depends on whether the curves intersect or do not intersect over this interval.

If the two curves f(x) and g(x) do not intersect over the interval [a, b] Here, we may look at three circumstances: when the region is above the x-axis, when it is below the x-axis, and when it crosses the x-axis.

Region above x-axis The red shaded area = =

y b

f(x)

b

∫a f ( x ) dx – ∫a g ( x ) dx

g(x)

b

0 a

∫a [ f ( x ) – g ( x ) ] dx .

x

b

Note: The lower function is subtracted from the higher function to ensure a positive answer.

Region below x-axis

y

Again, the lower function is subtracted from the higher function to ensure a positive answer. Red shaded area =

0

a

b

∫a [ f ( x ) – g ( x ) ] dx , as f(x) is above g(x)

b f(x)

x g(x)

over the interval [a, b]. y

Region crosses x-axis Shaded area =

f(x) g(x)

b

∫a [ f ( x ) – g ( x ) ] dx

0

a

b

x

WORKED Example 31

y

y = 2x + 1

a State the definite integral which describes the shaded area on the graph at right. b Find the area.

y=x

0

THINK WRITE a 1 State the two functions f(x) and g(x). a f(x) = 2x + 1 and g(x) = x Subtract the equation of the lower f(x) − g(x) = 2x + 1 − x 2 function from the equation of the upper =x+1 function and simplify. 3

Write as a definite integral between the given values of x.

2

Area =

∫0 [ f( x ) – g( x ) ]

Area =

∫0 ( x + 1 ) dx

2

2

x

393

Chapter 9 Integration

THINK

WRITE

b

b Area = [ 1--2- x 2 + x ] 02

1

Antidifferentiate.

2

Evaluate the integral.

Area = [ 1--2- ( 2 ) 2 + 2 ] – [ 1--2- ( 0 ) 2 + 0 ]

State the area.

Area = (2 + 2) − (0) Area = 4 The area is 4 square units.

3

WORKED Example 32

a Find the values of x where the functions y = x and y = x2 − 2 intersect. b Sketch the graphs on the same axes. (Check using a graphics calculator.) c Hence, find the area bound by the curves. THINK a 1 State the two functions. 2 Find where the curves intersect. 3 Solve for x.

WRITE a y = x and y = x2 − 2 For points of intersection: x = x2 − 2 x2 − x − 2 = 0 (x − 2)(x + 1) = 0 x = 2 or x = −1

b Find the key points of each function and sketch.

b For y = x, when x = 0, y = 0 when x = 2, y = 2 when x = −1, y = −1 Line passes through (0, 0), (2, 2) and (−1, −1) For y = x2 − 2, when x = 0, y = −2 Hence y-intercept is −2. Parabola also passes through (2, 2) and (−1, −1).

c

1

Define f(x) and g(x).

2

Write the area as a definite integral between the values of x at the points of intersection.

y

y = x2 – 2 (2, 2)

(–1, –1)

0

c Let f(x) = x and g(x) = x2 − 2 Area 2

=

∫–1 [ f ( x ) – g ( x ) ] dx

=

∫– 1 [ x – ( x

=

∫–1 ( x – x 2 + 2 ) dx

2

2

– 2 ) ] dx

2

Continued over page

y=x x

394

Mathematical Methods Units 3 and 4

THINK

WRITE

3

Antidifferentiate.

= [ 1--2- x 2 – 1--3- x 3 + 2x ] –21

4

Evaluate the integral.

= [ 1--2- ( 2 ) 2 – 1--3- ( 2 ) 3 + 2 ( 2 ) ] – [ 1--2- ( – 1 ) 2 – 1--3- ( – 1 ) 3 + 2 ( – 1 ) ] = (2 −

8 --3

+ 4) − ( 1--2- +

1 --3

− 2)

= (3 1--3- ) − (–1 1--6- ) = 3 1--3- + 1 1--6= 4 1--25

The area is 4 1--2- square units.

State the area.

If the two curves intersect over the interval [a, b] Where c1 and c2 are the values of x where f(x) and g(x) intersect over the interval [a, b], the area is found by considering the intervals [a, c1], [c1, c2] and [c2, b] separately. For each interval care must be taken to make sure the integrand is the higher function. Subtract the lower function. So the shaded area equals:

y g(x)

a c1 0

c2

x

b

f(x) c1

c2

b

∫a [ g( x ) – f( x ) ] d x + ∫c [ f( x ) – g( x ) ] d x + ∫c [ g( x ) – f( x ) ] d x 1

2

Therefore, when finding areas between two curves over an interval, it must be determined whether the curves intersect within that interval. If they do, the area is broken into sub-intervals as shown above. As with areas under curves, sketch graphs should be used to assist in finding areas between curves. If sketch graphs are not used, the absolute value of each integral, for each subinterval, should be taken to ensure the correct value is obtained.

WORKED Example 33

4 a Find the values of x where the graph of the functions f(x) = --- and g(x) = x intersect. x b Sketch the graphs on the same axes (check using a graphics calculator) and shade the region between the two curves and x = 1 and x = 3. c Find the area between f(x) and g(x) from x = 1 to x = 3. THINK

WRITE

a

4 a f(x) = --- , g(x) = x x

1

State the two functions.

2

Let f(x) = g(x) to find the values of x where the graphs intersect.

4 For points of intersection, x = --x

Chapter 9 Integration

THINK 3

WRITE x2 = 4 x2 − 4 = 0 (x − 2)(x + 2) = 0 x = −2 and x = 2

Solve for x.

b Sketch f(x) and g(x) on the same axes and shade the region between the two curves from x = 1 to x = 3.

b

y

f(x) = x4– g(x) = x x

01 2 3

c

395

c Area =

∫

2

 4--- – x dx +  1x

3

∫ 2  x – --x- dx 4

1

State the area as the sum of two integrals for the two subintervals.

2

Antidifferentiate.

= [ 4 log e x – 1--2- x 2 ] 12 + [ --21- x 2 – 4 log e x ] 23

3

Evaluate the two integrals.

= [ 4 log e 2 – 1--2- ( 2 ) 2 ] – [ 4 log e 1 – 1--2- ( 1 ) 2 ] + { [ 1--2- ( 3 ) 2 – 4 log e 3 ] – [ 1--2- ( 2 ) 2 – 4 log e 2 ] } = [ 4 log e 2 – 2 ] – [ 4 log e 1 – 1--2- ] + { [ 9--2- – 4 log e 3 ] – [ 2 – 4 log e 2 ] } = 4 log e 2 – 2 – 0 + 1--2- + 9--2- – 4log e 3 – 2 + 4log e 2

4

Add the two values.

5

State the area.

= 4 loge

4 --3

+1

The area is 4 loge

4 --3

+ 1 or approximately 2.151

square units.

Note: If worked example 33 was calculated without a graph, the area would be found by evaluating: 2

∫1  --x- – x dx 4

+

3

∫2  --x- – x dx 4

2

or

∫1  x – --x- dx 4

+

3

∫2  x – --x- dx 4

as without a graph it would not be known which function was above the other for either interval. To find the area between two curves using a graphics calculator: 1. Find the area between the upper curve and the x-axis, and record or store the answer. 2. Find the area between the lower curve and the x-axis, and record or store the answer. 3. Subtract the two answers, giving a positive value for the area.

396

Mathematical Methods Units 3 and 4

remember remember 1. If two curves f(x) and g(x) do not intersect over the interval [a, b] and f(x) > g(x) then the area enclosed by the two curves and the lines x = a and x = b is found by using the formula: b

∫a

b

∫a

f ( x ) dx –

b

∫a [ f ( x ) – g ( x ) ] dx

g ( x ) dx =

2. If two curves f(x) and g(x) intersect over the interval [a, b] it is necessary to find the points of intersection and hence find the area of each section because sometimes f(x) > g(x) and sometimes g(x) > f( x). 3. If sketch graphs are not used to determine which is the upper curve, then it is necessary to take the absolute value or positive value of each integral.

9H WORKED

Example

31a

Areas between two curves

1 State the definite integral which will find the shaded areas on each graph below. a

y

b

y = x2

c

y y = 2x

d

y

y

y = x2

y = 3x

y=x y=x+1 0

x

0 1

e y y = x3

f

1

y

0

x

2

g

y = ex

x 1 y = 4 – x2

y

y=x

h

y y = x2 – 5

y = 3x 01

0

2 x y = 8 – x2

0

–2

3

x

–1

2

x

–1

x

0

1 y = 9 – x2

0

x

1

y = –4 y = – ex

WORKED

Example

31b

2 Find each of the areas in question 1. 3 multiple choice Which one of the following does not equal the shaded area? 5

A

5

∫1 g ( x ) dx – ∫ 1 f ( x ) dx 5

5

C

∫1

E

∫5 [ f ( x ) – g ( x ) ] dx

1

f ( x ) dx –

∫1

g ( x ) dx

5

B

f(x)

∫ 1 g ( x ) dx + ∫5 f ( x ) dx 5

D

y

1

∫1

[ g ( x ) – f ( x ) ] dx

g(x) 0 1

5

x

397

Chapter 9 Integration

4 multiple choice The area bound by the curves f(x), g(x) and the lines x = −3 and x = 1 at right is equal to: –3

A

∫ –1[ f ( x ) – g ( x ) ] dx

C

∫ –3[ g ( x ) – f ( x ) ] dx

y g(x)

–1

B

∫ –3[ f ( x ) + g ( x ) ] dx

D

∫ –3[ f ( x ) – g ( x ) ] dx

–1

f(x)

–1

–4

0 –3

x

–1

–3

E

∫ –1[ f ( x ) + g ( x ) ] dx

5 multiple choice

y

g(x)

The shaded area at right is equal to: 4

A

∫0

f(x)

[ f ( x ) – g ( x ) ] dx

3

0

4

B

∫ 0 [ g ( x ) – f ( x ) ] dx + ∫ 3 [ f ( x ) – g ( x ) ] dx

C

∫ 0 [ g ( x ) – f ( x ) ] dx

4

3

E WORKED

Example

32

WORKED

Example

33

∫0

[ f ( x ) – g ( x ) ] dx +

3 4

x

3

D

∫0 [ f ( x ) – g ( x ) ] dx

4

∫3 [g( x) – f( x)]

6 In each of the following: i find the values of x where the functions intersect ii sketch the graphs on the same axes (check using a graphics calculator) iii hence, find the area bound by the curves. a y = 4x and y = x2 b y = 2x and y = 3 − x2 2 2 c y = x − 1 and y = 1 − x d y = x2 − 4 and y = 4 − x2 e y = (x + 1)2 and y = 1 − x2 f y = x and y = x2 7

i Find the values of x where the functions intersect. ii Sketch the graphs on the same axes. (Check using a graphics calculator.) iii Find the area between f(x) and g(x) giving an exact answer. a y = x3 and y = x b y = 3x2 and y = x3 + 2x

Use sketch graphs and a graphics calculator to assist in solving the following problems. 9 Find the area between the curve y = e x and the lines y = x, x = 1 and x = 3. x 10 Find the area between the curve y = x2 and the lines y = --- + 3, x = 1 and x = 3. 2

Math

cad

8 Find the area between the pairs of curves below, over the given interval. (Use a graph and graphics calculator to assist if necessary.) Area a y = x3, y = x2, x ∈ [−1, 1] b y = sin x, y = cos x, x ∈ [0, π] between c y = (x − 1)2, y = (x + 1)2, x ∈ [−1, 1] d y = x3 − 5x, y = 6 − 2x2, x ∈ [0, 3] two 1 curves e y = --- , y = 4x, x ∈ [ 1--4- , 1] f y = e x, y = e−x, x ∈ [0, 1] x π π g y = 2 cos x, y = x − --- , x ∈ [0, --- ] h y = e x, y = −e x, x ∈ [−2, 1] 2 2

398

Mathematical Methods Units 3 and 4

π 11 Calculate the area between the curves y = sin 2x and y = cos x from x = 0 to x = --- . 2 π 12 Calculate the area between the curves y = 3 − sin 2x and y = sin 2x from x = 0 to x = --- . 4 13 Find the exact area bound by the curves y = e x and y = 3 − 2e−x. 14 The graph at right shows the cross-section of a bricked archway. (All measurements are in metres.) a Find the x-intercepts of f(x). b Find the x-intercepts of g(x). c Find the cross-sectional area of the brickwork. 15 The diagram at right shows the outline of a window frame. If all measurements are in metres, what is the area of glass which fits into the frame?

y

y

f(x) = 4 – 1–4 x2

0 x g(x) = 3 – 1–3 x2

y = 2 1–2 – 2x2 y = 2x2

– 1–2

0

1– 2

x y

16 The diagram at right shows the side view of a concrete 5 bridge. (All measurements are in metres.) Find: x2 a the x-intercepts of the curve y = 4 – —– 100 b the length of the bridge x 0 c the area of the side of the bridge d the volume of concrete used to build the bridge if the bridge is 9 metres wide. 17 The cross-section of a road tunnel entrance is shown at right. (All measurements are in metres.) The shaded area is to be concreted. Find: a the exact area, above the entrance, which is to be concreted b the exact volume of concrete required to build this tunnel if it is 200 metres long.

y

0

x f(x) = 5 sin π–– 30

x

πx 18 A section of a river can be modelled by the equation y = 40 sin --------- , where x ∈ [0, 120] 120 and x is in metres. On the same model a proposed section of road obeys x the rule y = --- . The area bound by the road and the river is to have 5 one tree planted per 12 square metres. How many trees will be planted?

Chapter 9 Integration

399

Further applications of integration

9I

Further applications of integration

1 If f ′(x) = (2 − x)2 and the y-intercept of f(x) is 4--3- , find the rule for f(x). dy π 2 If ------ = 1 – 4 cos 2x and the y-intercept is 2, find the exact value of y when x = ------ . 12 dx 3 The rate of deflection from a hori- y (Metres) zontal position of a 3-metre diving (Metres) board when an 80-kg person is x x 0 Deflection metres from its fixed end is given by Board dy 2 ------ = – 0.03 ( x + 1 ) + 0.03, where dx y is the deflection in metres. a What is the deflection when x = 0? b Determine the equation which measures the deflection. c Hence, find the maximum deflection. 4 On any day the cost per item for a machine producing n items is given by dC ------- = 40 – 2e 0.01n , where n ∈ [0, 200] and C is the cost in dollars. dn a Use the rate to find the cost of producing the 100th item. b Express C as a function of n. c What is the total cost of producing the first 100 items? d Find the average cost of production for: i the first 100 items ii the second 100 items. 5 The rate of change of position (velocity) of a racing car travelling down a straight stretch ------ = t ( 16 – t ), where x is measured in metres and t in seconds. of road is given by dx dt a Find the velocity when: i t=0 ii t = 4. b Determine: i when the maximum velocity occurs ii the maximum velocity. dx c Sketch the graph of ------ against t for 0 ≤ t ≤ 16. dt d Find the area under the graph between t = 0 and t = 10. e What does this area represent?

400

Mathematical Methods Units 3 and 4

6 The rate at which water is pumped out of a dam, in L/min, t minutes after the pump is dV πt started is ------- = 5 + cos ------ . dt 40 a How much water is pumped out in the 40th minute? b Find the volume of water pumped out at any time, t, after the pump is started. c How much water is pumped out after 40 minutes? d Find the average rate at which water is pumped in the first hour. e How long would it take to fill a tank holding 1600 litres? 7 The rate of flow of water into a hot water system during a 12-hour period on a certain day is dV πt thought to be ------- = 10 + cos ----- , where V is in litres and t is the number of hours after 8 am. dt 2 dV a Sketch the graph of ------- against t. dt b Find the length of time for which the rate is above 10.5 L/h. c Find the volume of water that has flowed into the system between: i 8 am and 2 pm ii 3 pm and 8 pm. y (metres) 8 The roof of a stadium has the shape given by the function 20 f: [−25, 25] → R, f(x) = 20 − 0.024x2. The stadium is 75 metres long and its cross-section is 5 (metres) shown at right. x 0 –25 25 a Find the volume of the stadium. b The stadium is to have several airconditioners strategically placed around it. Each airconditioner can service a volume of 11250 m3. How many airconditioners are required? 9 The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full. 10 For any point P on the curve y = x3, prove that the area under the curve is one quarter of the area of the rectangle. y

y = x3 P

x

0

11 The arch of a concrete bridge has the shape of a parabola. It is 6 metres high and 8 metres long. y (metres) 7 6

–4

0

4 5

(metres) x

a Find the rule for the function corresponding to the arch of the bridge. b Find the area of the shaded region. c If the bridge is 10 metres wide, find the volume of concrete in the bridge.

Chapter 9 Integration

12 In the figure at right f(x) and g(x) intersect at O and B. a Show that the coordinate of B is (log e 2, 1). b Find the exact area of the region bound by f(x) and g(x). c Show that the sum of the areas under f(x) and g(x), from x = 0 to x = log e 2, is equal to the area of the rectangle OABC.

y

401

f(x) = ex – 1 B g(x) = –2e–x + 2

C

O

A

x

13 The population of kangaroos on an island is increasing at a rate given by P(t) = 12 log e (t + 1), where t is the number of years since 1 January 1955. a Find the rate of growth when t = 0, t = 5 and t = 40. b Sketch the graph of P(t). c Determine the inverse function P−1(t). d Use the inverse function to assist in finding the area under the graph of P(t) between t = 0 and t = 40. e What does this area represent?

Concrete chute

Concrete is poured from a mixer down a chute which y (cm) has a cross-sectional shape as shown. 5 The curved bottom has the shape given by the function f:[−10, 10] → R, f(x) = 0.12x2 − 12. 0 –10 a Find the area of the cross-section. Concrete can flow down the chute at 1.6 m/s. b What volume of concrete can be poured in one minute? –12 c How long does it take to empty a mixer holding 12 m3 of concrete?

(cm) x 10

402

Mathematical Methods Units 3 and 4

summary Antidifferentiation rules • The relationships between f(x) and

ax + c ax n + 1 --------------- + c n+1

a ax n (ax + b)n

( ax + b ) n + 1 ----------------------------- + c a(n + 1)

1 --x

log e x + c

1 --------------ax + b

1 --- log (ax + b) + c a

ex

ex + c 1 kx --- e + c k

e kx

•

are:

∫ f( x) d x

f(x)

•

∫ f ( x ) dx

sin ax

1 − --- cos ax + c a

cos ax

1 --- sin ax + c a

∫ [ f ( x ) ± g ( x ) ] dx = ∫ f ( x ) dx ± ∫ g ( x ) dx ∫ g ( x )dx = f ( x ) + c , where g(x) = f ′(x) dx

• •

∫ kf ( x ) dx = k ∫ f ( x ) dx ∫ f ( x ) dx is the indefinite integral

Definite integrals • The fundamental theorem of integral calculus: b

∫a f ( x ) dx = [ F ( x ) ]a = F(b) − F(a) b

where F(x) is an antiderivative of f(x).

b

•

∫a f ( x ) dx is the definite integral

•

∫a f ( x ) dx = ∫a f ( x ) dx + ∫c f ( x ) dx , a < c < b

•

∫a kf ( x ) dx = k ∫a f ( x ) dx

•

∫a [ f ( x ) ± g ( x ) ] dx = ∫a f ( x ) dx ± ∫a g ( x ) dx

•

∫a f ( x ) dx = –∫b f ( x ) dx

b

c

b

b

b

b

b

b

a

b

403

Chapter 9 Integration

Approximating areas under curves • An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles or trapeziums which are all the same width. The approximation is found by finding the sum of all the areas of the rectangles or trapeziums. • Lower rectangle approximation ≤ actual area ≤ upper rectangle approximation • Trapezoidal approximation = lower rectangle approximation + upper rectangle approximation --------------------------------------------------------------------------------------------------------------------------------------------------------2 • Trapezoidal rule is b h f ( x ) dx ≅ --- [ f ( a ) + 2f ( a + h ) + 2f ( a + 2h ) + … + 2f ( b – h ) + f ( b ) ] a 2 where h is the interval width.

∫

y

Area under curves • Area =

y = f(x)

b

∫a f ( x ) dx , if f(x) > 0 for x ∈ [a, b] 0

a y

y = f(x) a

• Area = −

• Area = =

b

∫a

f ( x ) dx , if f(x) < 0 for x ∈ [a, b]

b

c

b

c

x

b

b x

0

∫c f ( x ) dx – ∫a f ( x ) dx

y

y = f(x)

∫c f ( x ) dx + ∫a f ( x ) dx , if f(x) > 0 for x ∈ [c, b]

and f(x) < 0 for x ∈ [a, c]

A1 0 A 2

a

c

b

Area between curves y

• Area =

b

∫a

f(x)

[ f ( x ) – g ( x ) ] dx , if f(x) > g(x) for x ∈ [a, b] g(x) 0 a

• Area =

c

∫a

[ g ( x ) – f ( x ) ] dx +

b

∫c

[ f ( x ) – g ( x ) ] dx ,

y

x

b g(x) f(x)

if g(x) > f (x) for x ∈ [a, c] and f (x) > g(x) for x ∈ [c, b] a 0

c

b

x

x

404

Mathematical Methods Units 3 and 4

CHAPTER review Multiple choice

9A

9A

1 1 The antiderivative of 4x 3 – ----------- is: 1–x A x4 − log e (1 − x) + c B x4 + log e (1 − x) + c 1 E x 4 – ------------------2- + c D 16x4 + log e (1 − x) + c (1 – x) 2 The indefinite integral ( 5x – 4 ) 4 dx is equal to: A 25(5x − 4) + c 5

D

9A

9B

∫

1 --- (5x 5

− 4)5 + c

1 ------ (5x 25

C (5x − 4)5 + c

− 4)5 + c

3 An antiderivative of 2(3x + 4)−4 is: A − 2--3- (3x + 4)−3

B − 2--3- (3x + 4)−3 + 5

D − 2--9- (3x + 4)−5

E −8(3x + 4)−3

4 The antiderivative of 6e−3x is: A −2e−3x + c B −3e−3x + c

C − 2--9- (3x + 4)−3

C −18e−3x + c

E − 1--2- e−3x + c

x 5 The indefinite integral  cos --- – 3 sin 3x dx is equal to:   3

∫

x A sin --- + cos 3x + c 3 x D 3 sin --- + cos 3x + c 3

9B

B 5(5x − 4)5 + c E

D −2e−3x + 1 + c

9B

C 16x4 − log e (1 − x) + c

x --- + 3 cos 3x + c 3 x E 3 sin --- − cos 3x + c 3

B

1 --- sin 3

x C −3 sin --- + cos 3x + c 3

6 An antiderivative of x3 + sin 4x + e4x is: A 4(x4 − cos 4x + e4x) D

1 4 --- (x 4

− cos x + e4x)

B

1 4 --- x 4

+ cos 4x + 1--4- e4x

E

1 4 --- (x 4

C

1 --4

x4 − 4 cos 4x + 1--4- e4x

− cos 4x + e4x)

9B

7 If f(x) has a stationary point at (0, 3) and f ′(x) = e x + k, where k is a constant, then f(x) is: A e x − 2x + 2 B −e x − x + 2 C ex − x + 2 –x x D e +x+2 E e + 2x + 1

9C

8 If the derivative of (x − x2)8 is 8(1 − 2x)(x − x2)7 then an antiderivative of 24(1 − 2x)(x − x2)7 is: A 2(x − x2)8 B 3(x − x2)8 C 1--2- (x − x2)8 1 2 8 2 8 D --3- (x − x ) E 8(x − x )

405

Chapter 9 Integration

3

9 If the derivative of e x 3

A 3ex D

+ 3x

1 x3 + 3x --- e 3

+ 3x

+c +c

is 3(x2 + 1)ex + 3x, then the antiderivative of (x2 + 1)e x 2 1 B ---------------------C e3x + 3 + c +c 2 3( x + 1) E undefined 3

3

+ 3x

is:

9C

x – 2x 10 If the derivative of log e (5 − x2) is -------------2- then the antiderivative of -------------2- is: 5–x 5–x A − 1--2- log e (5 − x2) + c B −2 log e (5 − x2) + c C 1--2- log e (5 − x2) + c 2 D 2 log e (5 − x ) + c E undefined

9C

y (3.5, 9) (3, 6) (2.5, 4) (2, 3)

11 The approximation for the area under the graph at right from x = 2 to x = 4, using the ‘lower rectangles’ is: A 22 sq. units D 10 sq. units

B 14 sq. units E 20 sq. units

C 11 sq. units 0

B 21 sq. units E 10 sq. units

C 23 sq. units

13 Using the trapezoidal rule the area under the curve at right from x = 1 to x = 4 is approximately equal to: B 16 sq. units E 19 sq. units

14 The expression A 2

∫0 ( 3

C 22 sq. units

A 3e4 − e−4

B 2e4 − e−4

16 The exact value of A −3

π

D 20

1 2 3 4 x

E 16

2

∫–2 ( 4e2 x – 2e–2 x ) dx is:

C e4 − 2e−4

9E

D e4 + e−4

E e4 − e−4

D 3 3

E 3 2

x

9E

C –3 3

17 The shaded area on the graph at right is equal to: A 12 sq. units D 4 sq. units

9D

∫0  –2 cos --3- dx is:

B 3

B 16 sq. units E 8 sq. units

x

9E

C −2 1--2-

15 The exact value of the definite integral

–1 0

(4, 10) (3, 7) (2, 5) (1, 4)

x – x ) dx is equal to:

B 8

9D

y

0 4

4 x

(–5, 8) (–4, 6) (–3, 5) (–2, 4)

–5

A 26 sq. units D 17 sq. units

3

2

y

12 The area under the graph at right from x = −5 to x = −1 can be approximated by the area of the ‘upper rectangles’ and is equal to: A 20 sq. units D 11 1--2- sq. units

9D

y y = (x – 2)3

C 10 sq. units 0

2

4 x

9F

406 9F

9G

Mathematical Methods Units 3 and 4

18 The shaded area on the graph at right is:

y

B −20 sq. units E −18 sq. units

A 20 sq. units D 16 sq. units

C −16 sq. units

0

–2 –1

y = –1 – 3x2

19 The area bound by the curve on the graph at right and the x-axis is equal to: A

5 20 ----12

sq. units

B

5 - sq. units D −10 ----12

1 21 ----12

sq. units

x

2

5 10 ----12

C

y y = x(x + 2)(x – 3)

sq. units

7 - sq. units E 20 ----12

x

0

Questions 20 to 22 apply to the curve with equation f(x) = e x − 1. 20 The graph of f(x) is best represented by: A

y

B f(x)

y

f(x)

C

y

f(x)

2 0

x

0

D

y

–1

x

0

f(x) 0

x

E

y

f(x) x

–1 0 –1

x

–1

9G

21 The area bound by the graph of f(x), the x-axis and the line x = 2 is equal to:

9G

22 The area bound by the graph of f (x), the y-axis and the line y = e2 − 1 is equal to:

A e2 − 1

A e2 − 5

B e2 − 2

B e2 − 3

C e2 + 1

C e2 + 2

D e2 + 1

D e2 + 2

E e2 − 3

E 5 − e2

Use the graph at right to answer questions 23 and 24. y

9H

23 The two graphs intersect where x is equal to:

9H

24 The area bound by the two graphs is equal to:

A 1 and −3 B −1 and 3 C 1 and 2

9A 9B

D −1 and −2 E 1 and 3

A 10 2--3- sq. units

B 7 1--3- sq. units

D 11 1--3- sq. units

E 6 2--3- sq. units

Short answer

y = x2

y = 2x + 3 0

x

C −7 1--3- sq. units

3x 3 – 2x 2 1 Find the equation of the curve f(x) if it passes through (1, −3) and f′(x) = ----------------------- . x dy πx 2 A particular curve has ------ = cos ------ + k , where k is a constant, and it has a stationary point dx 4 (2, 1). Find: a the value of k b the equation of the curve c the value of y when x = 6.

Chapter 9 Integration

407

dy 3 If y = sin (x2 + 2x), find ------ and hence antidifferentiate (x + 1) cos (x2 + 2x). dx

9C

4 Apply the trapezoidal method to find the area between y = e2x − 1 and the x-axis, from x = 0 to x = 4, using intervals of width 1 unit. (Give an exact answer.)

9D

5 Calculate an approximation for the area under the curve y = log e x from x = 2 to x = 4, using interval widths of 0.5 units.

9D

6 Evaluate each of the following definite integrals.

9E

a

∫

0

9 – ----------------------4 dx – 1 ( 2x + 3 )

7 Given that

b

∫

2π -----3

π --3

cos 2x dx

k

∫0 ( 4x – 5 ) dx = –2 , find two possible values for k.

1 8 a Sketch the graph of the function f(x) = ----------- . x–2 b Find the exact area between the graph of f(x), the x-axis and the lines x = 3 and x = 6. 9 Find the area bound by the curve g(x) = (4 − x)(6 + x) and the x-axis.

π 10 Calculate the area between the curve y = 2 cos x and the lines y = −x, x = 0 and x = --- . 2

Analysis 1 From past records it has been found that the cost rate of maintaining a certain car is dC ------- = 75t 2 + 50t + 800 , where C is the accumulated cost in dollars and t is the time in years dt since the car was first used. Find: a the initial maintenance cost, b C as a function of t c the total maintenance cost during the first 5 years of use of the car d the total maintenance cost from 3 to 5 years e the maintenance cost for the second year. 2 Over a 24-hour period on a particular autumn day, starting at 12 am, the rate of change of the πt dT 5π temperature for Melbourne was approximately ------- = – ------ cos ------ , where T is the 12 12 dt temperature in °C and t is the number of hours since midnight when the temperature was 10°C. Find: a the temperature at any time, t b whether the temperature reaches 17°C at any time during the day c the maximum temperature and the time at which it occurs d the minimum temperature and the time at which it occurs e the temperature at i 2 am ii 3 pm f the time when the temperature first reaches 14.33°C.

9E 9F 9G 9H

408

Mathematical Methods Units 3 and 4 x ---

3 The diagram at right shows part of the curve with equation y = e 2 . y Find: normal y=e B a the coordinate of point A b the equation of the normal to the curve at point A A C c the coordinate of point B d the coordinate of point C x 0 2 e the area bound by the curve and the lines AB and BC. –x 2

y

4 a Find the derivative of x log e x. 1 b Hence, find an antiderivative of log e x. e–2 The cross-section of a platform is shown at right. (All 0 e x 1 measurements are in metres.) f(x) = log ex c Find the height of the platform. d Find the cross-sectional area of the platform. e Find the volume of concrete required to build this platform if it is 20 metres long.

CHAPTER

test yourself

9