MATHEMATICS I (Engineering) · MATHEMATICS I (Engineering) STUDY GUIDE 2 for MAT1581 L E Greyling...

MATHEMATICS I

(Engineering) STUDY GUIDE 2 for MAT1581

L E Greyling Department of Mathematical Sciences UNIVERSITY OF SOUTH AFRICA PRETORIA

© UNISA 2016 First edition 2004 Second edition 2005 Third edition 2017 All rights reserved Printed and published by the University of South Africa Muckleneuk, Pretoria MAT1581 Layout done by the Department

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CONTENTS PAGE Introduction ii

STUDY GUIDE 2

MODULE 6

Differentiation

Learning unit 1 Functional notation 4 Learning unit 2 Limits 11 Learning unit 3 The derivative 26 Learning unit 4 Standard forms 34 Learning unit 5 Rules of differentiation I 41 Learning unit 6 Rules of differentiation II 51 Learning unit 7 Higher order derivatives 70 Learning unit 8 Applications I 79 Learning unit 9 Applications II: maxima and

minima 88

Post-test 109 MODULE 7 Integration

Learning unit 1 Reverse of differentiation I 137 Learning unit 2 Reverse of differentiation II 144 Learning unit 3 Method of substitution 152 Learning unit 4 Standard integrals 161 Learning unit 5 Partial fractions 172 Learning unit 6 Trigonometric functions 178 Learning unit 7 The definite integral 187 Learning unit 8 Areas 193 Post-test 210

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INTRODUCTION:

MATHEMATICS I (Engineering) Study guide 1 dealt with topics from pre-calculus. You are now ready to start your study of calculus. Read the introduction in study guide 1 to refresh your memory. You need to keep refering to the formula sheets and for your convenience it is reprinted at the beginning of this study guide.

USEFUL INFORMATION

MATHEMATICAL SYMBOLS + plus minus plus or minus multiply by multiply by divide by = is equal to is identically equal to is approximately equal to is not equal to is greater than is greater than or equal to is less than is less than or equal to n! factorial n = 1 2 3 ….. n k modulus of k, that is the size of k

irrespective of the sign is a member of set set of natural numbers set of integers set of real numbers set of rational numbers

therefore infinity e base of natural logarithms (2,718…) ln natural logarithm log logarithm to base 10 sum of terms limn

limiting value as n

integral dy

dx derivative of y with respect to x

Formula sheets The following pages contain the information sheets and table of integrals that will be included with the examination paper.

(iii)

FORMULA SHEET ALGEBRA Laws of indices

n

nn

n

n

n

n

n mn

m

mnmnnm

nm

n

m

nmnm

b

a

b

a

baab

a

aa

aa

aa

aaa

aa

a

aaa

.8

.7

1.6

1and

1.5

.4

.3

.2

.1

0

Logarithms

Definitions If xay then yx alog

If xey then ynx

Laws

log n

1. log log log

2. log log log

3. log log

log4. log

log

5. a

n

ba

b

f f

A B A B

AA B

B

A n A

AA

a

a f e f

Factors

2233

2233

babababa

babababa

Partial fractions

dx

C

cbxax

BAx

dxcbxax

xf

bx

D

ax

C

ax

B

ax

A

bxax

xf

cx

C

bx

B

ax

A

cxbxax

xf

22

323

Quadratic formula

a

acbbx

cbxax

2

4then

0If

2

2

DETERMINANTS

223132211323313321122332332211

3231

222113

3331

232112

3332

232211

333231

232221

131211

aaaaaaaaaaaaaaa

aaaa

aaaaa

aaaaa

aaaaaaaaaa

(iv)

SERIES Binomial theorem

11and

...!3

21

!2

111

and

..3

21

2

1

32

33221

x

xnnn

xnn

nxx

ab

..ba!

nnnba

!

nnbnaaba

n

nnnnn

Maclaurin’s theorem

11

32

!1

0

!3

0

!2

0

!1

00 n

n

xn

fx

fx

fx

ffxf

Taylor’s theorem

afn

haf

haf

hafhaf

axn

afax

afax

afax

afafxf

nn

nn

112

11

32

!1!2!1

!1!3!2!1

COMPLEX NUMBERS

212

1

2

1

212121

22

2

:Division.6

:tionMultiplica.5

andthen,If.4

:nSubtractio.3

:Addition.2

tanarg:Argument

:Modulus

1where

,sincos.1

r

r

z

z

rrzz

qnpmjqpjnm

dbjcajdcjba

dbjcajdcjba

a

barcz

bazr

j

rerjrbjaz j

1

1 1

7. De Moivre's theorem

cos sin

8. has distinct roots:

360with 0, 1, 2, , 1

9. cos sin

cos and sin

10. cos sin

11.

n n n

n

n n

j

j j

a jb a

j

r r n r n j n

z n

kz r k n

n

re r j

re r re r

e e b j b

n re n r j

(v)

GEOMETRY

1. Straight line

11 xxmyy

cmxy

Perpendiculars, then 2

1

1

mm

2. Angle between two lines

21

21

1tan

mm

mm

3. Circle

222

222

rkyhx

ryx

4. Parabola cbxaxy 2

axis at a

bx

2

5. Ellipse

12

2

2

2

b

y

a

x

6. Hyperbola

axis- round1

axis- round1

2

2

2

2

2

2

2

2

yb

y

a

x

xb

y

a

x

kxy

MENSURATION 1. Circle: ( in radians)

2

2

2

Area

Circumference 2

Arc length

1 1Sector area

2 21

Segment area sin2

r

r

r

r r

r

2. Ellipse

ba

ab

nceCircumfere

Area

3. Cylinder

2

2

22area Surface

Volume

rrh

hr

4. Pyramid

height base area3

1Volume

5. Cone

r

hr

surfaceCurved3

1Volume 2

6. Sphere

3

2

3

4

4

rV

rA

7. Trapezoidal rule

0 1 11

2 22 n n

b af x f x f x f x

n

8. Simpson’s rule

0 1 2 3

4 2 1

1[ 4 2 4

3

2 2 4 ]n n n

b af x f x f x f x

n

f x f x f x f x

9. Prismoidal rule

1 2 1n nb a

f m f m f m f mn

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HYPERBOLIC FUNCTIONS Definitions

xx

xx

xx

xx

ee

eex

eex

eex

tanh

2cosh

2sinh

Identities

x

x

xxx

xxx

xx

xx

xx

xx

xx

2

2

22

2

2

22

22

22

sinh21

1cosh2

sinhcosh2cosh

coshsinh22sinh

12cosh2

1cosh

12cosh2

1sinh

cosech1coth

sechtanh1

1sinhcosh

TRIGONOMETRY Compound angle addition and subtraction formulae sin(A + B) = sin A cos B + cos A sin B sin(A - B) = sin A cos B - cos A sin B cos(A + B) = cos A cos B - sin A sin B cos(A - B) = cos A cos B + sin A sin B

BA

BABA

BA

BABA

tantan1

tantantan

tantan1

tantantan

Double angles sin 2A = 2 sin A cos A cos 2A = cos2A – sin2A = 2cos2A - 1 = 1 - 2sin2A sin2 A = ½(1 - cos 2A) cos2 A = ½(1 + cos 2A)

A

AA

2tan1

tan22tan

Products of sines and cosines into sums or differences sin A cos B = ½(sin (A + B) + sin (A - B)) cos A sin B = ½(sin (A + B) - sin (A - B)) cos A cos B = ½(cos (A + B) + cos (A - B)) sin A sin B = -½(cos (A + B) - cos (A - B)) Sums or differences of sines and cosines into products

2sin

2sin2coscos

2cos

2cos2coscos

2sin

2cos2sinsin

2cos

2sin2sinsin

yxyxyx

yxyxyx

yxyxyx

yxyxyx

TRIGONOMETRY Identities

cos

sintan

tan- = )(-tan

cos = )(- cos

sin - = )sin(-

cosec = 1 +cot

sec = tan+ 1

1 cos sin

22

22

22

(vii)

DIFFERENTIATION

h 0

1

2

1

1. lim

2. 0

3.

4. . . ' . '

. ' . '5.

6. ( ) ( ) . '( )

7. . .

n n

n n

f x h f xdy

dx hd

kdxd

ax anxdxd

f g f g g fdxd f g f f g

dx g g

df x n f x f x

dxdy dy du dv

dx du dv dx

8. Parametric equations

2

2

dydy dt

dxdxdt

d dyd y dt dx

dxdxdt

9. Maximum/minimum For turning points: f '(x) = 0

Let x = a be a solution for the above If f '(a) > 0, then a minimum If f '(a) < 0, then a maximum For points of inflection: f " (x) = 0 Let x = b be a solution for the above

Test for inflection: f (b h) and f(b + h) Change sign or f '"(b) if f '"(b) exists

'( )110. sin ( )21 ( )

'( )111. cos ( )21 ( )

'( )112. tan ( )21 ( )

'( )113. cot ( )21 ( )

'( )114. sec ( )2( ) 1

'( )115. cosec ( )2( ) 1

'( )116. sinh ( )

d f xf x

dxf x

d f xf x

dxf x

d f xf x

dx f x

d f xf x

dx f x

d f xf x

dxf x f x

d f xf x

dxf x f x

d f xf x

dxf

2( ) 1

'( )117. cosh ( )2( ) 1

'( )118. tanh ( )21 ( )

'( )119. coth ( )2( ) 1

'( )120. sech ( )21 ( )

'( )121. cosech ( )2( ) 1

22. Increments: . . .

x

d f xf x

dxf x

d f xf x

dx f x

d f xf x

dx f x

d f xf x

dxf x f x

d f xf x

dxf x f x

z z zz x y w

x y w

23. Rate of change:

. . .z z zdz dx dy dw

dt x dt y dt w dt

INTEGRATION

b

a

b

a

b

a

dxyb-a

dxyb-a

F(aF(b)dxf(x)vduuv-udv

22 1)R.M.S.(.4

1= Mean value.3

).2:partsBy.1

(viii)

TABLE OF INTEGRALS

1

1

1 11

2 ' 11

'3

4 '

5 '

6 ' sin cos

7 ' cos sin

8 '

(n )n

nn

f(x) f(x)

f(x)f(x)

a x. ax dx c, n

n

f(x). f(x) .f (x) dx c, n

n

f (x). dx n f(x) c

f(x)

. f (x).e dx e c

a. f (x).a dx c

n a

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f(x) c

. f (x)

2

2

tan sec

9 ' cot sin

10 ' sec sec tan

11 ' cosec osec cot

12 ' sec tan

13 ' cosec cot

14

. f(x) dx n f(x) c

. f (x). f(x) dx n f(x) c

. f (x). f(x) dx n f(x) f(x) c

. f (x). f(x) dx n c f(x) f(x) c

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f(x) c

' sec tan sec

15 ' cosec cot cosec

. f (x). f(x). f(x) dx f(x) c

. f (x). f(x). f(x)dx f(x) c

1 MAT1581 Mathematics 1 (Engineering)

M O D U L E 6

DIFFERENTIATION CONTENTS Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject which falls into two parts: differentiation (module 6) and integration (module 7).

PAGE

LEARNING UNIT 1 FUNCTIONAL NOTATION 4

1. FUNCTIONS .......................................................................................................... 5 1.1 Dependent and independent variables .................................................................... 5 1.2 Functional notation ................................................................................................. 5 2. RESPONSES TO ACTIVITY ................................................................................ 8

LEARNING UNIT 2 LIMITS 11

1. INTRODUCTION ................................................................................................ 12 2. THE TANGENT QUESTION .............................................................................. 12 3. WHAT IS MEANT BY A LIMIT? ...................................................................... 14 4. TECHNIQUES FOR FINDING LIMITS ............................................................. 15 4.1 When the limit of the denominator of a quotient is 0 ........................................... 18 4.2 Limits at infinity ................................................................................................... 19 5. ONE-SIDED LIMITS AND CURVES ................................................................ 21 6. RESPONSES TO ACTIVITIES ........................................................................... 23 6.1 Activity 1 .............................................................................................................. 23 6.2 Activity 2 .............................................................................................................. 24 6.3 Activity 3 .............................................................................................................. 25

LEARNING UNIT 3 THE DERIVATIVE 26

1. THE TANGENT QUESTION AND THE DERIVATIVE .................................. 27 2. DIFFERENTIATION FROM FIRST PRINCIPLES ........................................... 29 3. RESPONSES TO ACTIVITY .............................................................................. 31

LEARNING UNIT 4 STANDARD FORMS 34

1. STANDARD FORMS .......................................................................................... 35 2. RESPONSES TO ACTIVITIES ........................................................................... 39 2.1 Activity 1 .............................................................................................................. 39 2.2 Activity 2 .............................................................................................................. 39


M O D U L E 6

LEARNING UNIT 5 RULES OF DIFFERENTIATION I 41

1. DERIVATIVES OF COMBINED EXPRESSIONS ............................................ 42 1.1 Constant times a function ...................................................................................... 42 1.2 Sums and differences ............................................................................................ 42 1.3 Product rule ........................................................................................................... 45 1.4 Quotient rule ......................................................................................................... 46 2. RESPONSES TO ACTIVITIES ........................................................................... 47 2.1 Activity 1 .............................................................................................................. 47 2.2 Activity 2 .............................................................................................................. 47 2.3 Activity 3 .............................................................................................................. 48 2.4 Activity 4 .............................................................................................................. 49 2.5 Activity 5 .............................................................................................................. 49

LEARNING UNIT 6 RULES OF DIFFERENTIATION II 51

1. DERIVATIVES OF COMPOSITE FUNCTIONS .............................................. 52 1.1 Function-of-a-function or chain rule ..................................................................... 52 1.2 General standard forms ......................................................................................... 61 2. RESPONSES TO ACTIVITIES ........................................................................... 65 2.1 Activity 1 .............................................................................................................. 65 2.2 Activity 2 .............................................................................................................. 66 2.3 Activity 3 .............................................................................................................. 66 2.4 Activity 4 .............................................................................................................. 66 2.5 Activity 5 .............................................................................................................. 68

LEARNING UNIT 7 HIGHER ORDER DERIVATIVES 70

1. HIGHER ORDER DERIVATIVES ..................................................................... 71 2. RESPONSES TO ACTIVITY .............................................................................. 77

LEARNING UNIT 8 APPLICATIONS I 79

1. L’HOSPITAL’S RULE ........................................................................................ 80 2. THE GRADIENT OF A CURVE ......................................................................... 81 2.1 The equation of a tangent to a curve ..................................................................... 81 2.2 The equation of a normal to a curve ..................................................................... 83 3. RATE OF CHANGE ............................................................................................ 84 3.1 Velocity ................................................................................................................. 84 3.2 Acceleration .......................................................................................................... 84 4. RESPONSES TO ACTIVITIES ........................................................................... 86 4.1 Activity 1 .............................................................................................................. 86 4.2 Activity 2 .............................................................................................................. 87 4.3 Activity 3 .............................................................................................................. 87

MAT1581 3 Mathematics 1 (Engineering)

M O D U L E 6

LEARNING UNIT 9 APPLICATIONS II: MAXIMA AND MINIMA 88

1. MAXIMA AND MINIMA ................................................................................... 89 1.1 Definitions............................................................................................................. 89 1.2 Maximum value .................................................................................................... 90 1.3 Minimum value ..................................................................................................... 90 2. DERIVED CURVES ............................................................................................ 91 3. PRACTICAL APPLICATIONS ........................................................................... 99 4. RESPONSES TO ACTIVITIES ......................................................................... 103 4.1 Activity 1 ............................................................................................................ 103 4.2 Activity 2 ............................................................................................................ 107

POST-TEST 109

POST-TEST SOLUTIONS 113


DIFFERENTIATION Functional notation

CONTENTS PAGE

1. FUNCTIONS .......................................................................................................... 5 1.1 Dependent and independent variables .................................................................... 5 1.2 Functional notation ................................................................................................. 5 2. RESPONSES TO ACTIVITY ................................................................................ 8

MODULE 6

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to use functional notation determine the value of a function interpret functional notation

Module 6 Learning unit 1 DIFFERENTIATION: Functional notation


1. FUNCTIONS

A function is a special type of relation. It is a relation in which each element of the domain corresponds to only one element of the range. However, the reverse is not true, which implies that each element of the range does not necessarily correspond to only one element of the domain.

1.1 Dependent and independent variables

An equation involving two variable quantities has the property that when a value is assigned to one of the variable quantities, the other is determined.

For example, the letters x and y indicate the variable quantities in 3 23 5y x x . If we assign a value to x in the equation, we can determine y. Say we assign the value 0 to x; we can determine the value of y as 5. Because x is the value which is assigned, it is called the independent variable. An arbitrary value can be assigned to x. Because the value of y is dependent on the value assigned to x, y is called the dependent variable.

1.2 Functional notation

The functional notation is mathematical shorthand which is extremely convenient to use. It gives an expression of an identification tag within a certain area of discussion and specifies the variables involved. For example, instead of writing out expressions

such as 4 2 4 3 2 75 3 4 and 3 5 4x x x x x y we can identify them in the following way:

4 2

4 3 2 7

5 3 4 and

, = 3 5 4

f x x x

G x y x x x y

Now f and G are the tags, which identify the functions in our discussion, and the x and y in brackets indicate the independent variables. Once identified, we refer to f(x) and G(x,y) in our discussion or problem-solving instead of writing the expressions in full. This notation is also useful when we work with equations.

For example, if 3 23 5y x x and we let 3 23 5f x x x , then we can say

y f x . This clearly indicates x as the independent variable and we say y is a

function of x. If we need to discuss more than one expression or equality in the same context, then we must have several tags so that each one is clearly distinguished. Other letters in common use as identification tags are F, G, g and the Greek letters and . We often need to know the value of a function for an assigned value of the independent variable. We can use a graph or calculate the value if the function is known.



Example 1

If y f x figure 1 shows the meaning of f (2).

The length of the ordinate at the point x = 2 represents the value f (2).

Figure 1

In general f (x) is the length of the ordinate at any point x.

Example 2

If 2( ) 1f x x , find the value of a) f (1) b) f (3) and c) f (a)

b) 21 1 1 1 1 2f

c) 23 3 1 9 1 10f

d) 2 21 1f a a a

Example 3

Given that 2 4 7

2

x xG x

x

find 0 , 10 and G G G x h .

2

2

0 4 0 70

0 2

73,5

2

10 4 10 710

10 2

100 40 7

12

14712,25

12

G

G



2

2 2

4 7

2

2 4 4 7

2

x h x hG x h

x h

x xh h x h

x h

Example 4

If angular displacement is given by 2½wt at , where w and a are constants and t is the time in seconds, evaluate and interpret = f(0) and = f(10). Solution In our shorthand notation, since is a function of the time t, we have f t .

212

Then 0

0 0

0

f

w a

Thus the displacement is 0 when the time is counted 0, which is usually at the beginning of the movement.

Further 21210 10 10

10 50

f w a

w a

that is, the displacement is 10 50w a units when the time has moved 10 seconds from the start of the counting.

ACTIVITY 1

1. Given 2( ) 5 2 1f x x x determine

0 , 5 , 7 , 1 and f f f f f x h .

2.

2If 3 5 8 find

a) 1

b) 3 1

c) 2 1

x x x

3.

3If 6 , find

a) 3 1 4 1

b)

c)

F x x

F F

F x h F x

F h

h



4.

2If 2 , find expressions for

a)

b)

c)

g x x

g x h

g x h g x

g x h g x

h

5. Consider the following two functions: f(x) = 2x2 + 3x and h(x) = 3x3 x + 4.

Calculate the value of 22 2f x h x .

6. Consider the functions 2( ) 4 2 and ( ) 3f t t t g s s . Find the following values: a) (2)f b) (9)g c) 2 (1) 3 (2)f g

7. Given that 3 3 1 and 2 3f x x x g t t ,

determine the following values:

a) 3

b) 3

c) 1

d) ( 2)

f

f

g a g a

f g

Remember to check the response on page 8.

2. RESPONSES TO ACTIVITY

Activity 1

1. 2(0) 5 0 2 0 1

1

f

2

2

(5) 5 5 2 5 1

125 10 1

116

(7) 5 7 2 7 1

245 14 1

232

f

f

2( 1) 5 1 2 1 1

5 2 1

8

f



2

2 2

2 2

( ) 5 2 1

5 2 2 2 1

5 10 5 2 2 1

f x h x h x h

x xh h x h

x xh h x h

2 2 3

2

2 2

2

2. a) 10

b) 0

c) 24

3. a) 42

b) 18 18 6

c) 6

4. a) 2 4 2

b) 4 2

c) 4 2

x h xh h

h

x hx h

hx h

x h

5.

22 2 3

4 3 2 3

4 3 2 3

4 3 2

2[ ( )] - 2[ ( )]=2 2 + 3 2 3 + 4

2 4 12 9 6 2 8

8 24 18 6 2 8

8 18 18 2 8

f x h x x x x x

x x x x x

x x x x x

x x x x

6. a) (2) = 2² + 4(2) + 2

= 4 + 8 + 2

= 14

f

b) (9) = 9 + 3.

= 12

c) 2 (1) + 3 (2) = 2[(1)² + 4(1) + 2] + 3[2 + 3]

= 2[7] + 3[5]

= 29

g

f g

7. 3a) 3 3 3 3 1

3 3 3 3 1

1

f



3b) 3 3 3 3 1

3 3 3 3 1

1

f

c) 1

2 1 3 2 3

2 2 3 2 3

2

g a g a

a a

a a

3

d) First calculate 2

2 2 2 3

4 3

7

Thus ( 2) 7

7 3 7 1

321

g

g

f g f

This learning unit 1 focused on functional notation and you should be able to use functional notation determine the value of a function interpret functional notation

The next learning unit explores limits.


DIFFERENTIATION Limits

CONTENTS PAGE

1. INTRODUCTION ................................................................................................ 12 2. THE TANGENT QUESTION .............................................................................. 12 3. WHAT IS MEANT BY A LIMIT? ...................................................................... 14 4. TECHNIQUES FOR FINDING LIMITS ............................................................. 15 4.1 When the limit of the denominator of a quotient is 0 ........................................... 18 4.2 Limits at infinity ................................................................................................... 19 5. ONE-SIDED LIMITS AND CURVES ................................................................ 21 6. RESPONSES TO ACTIVITIES ........................................................................... 23 6.1 Activity 1 .............................................................................................................. 23 6.2 Activity 2 .............................................................................................................. 24 6.3 Activity 3 .............................................................................................................. 25

MODULE 6

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to explain what is meant by a limit find the value of a limit give examples of one-sided limits

Module 6 Learning unit 2 DIFFERENTIATION: Limits


1. INTRODUCTION

It is often important to know how quickly a quantity is changing, for example the rate at which the speed of a car is increasing or decreasing, the rate at which the temperature of a chemical is rising in a tank and the rate at which the currency is fluctuating. Differentiation focuses on analysing the rate at which a function is changing in a situation. Graphically differential calculus solves the tangent question.

2. THE TANGENT QUESTION

In module 5, unit 1, we studied the slope of a straight line. We found that if

1 1, andx y 2 2,x y are two points on a line, then the slope of the line is given by

2 1

2 1

y ym

x x

.

Consider the curve in figure 1.

Figure 1

This is not a straight line, but it is the curve of a function. The slope of a straight line is the same at every point on the line. The curve in figure 1 gives the impression that it gets steeper as x increases. We might expect that the slope of a non-linear curve would be different at different points on the curve. We would like a way to measure the steepness or slope of such a curve at any specific point on the curve. The slope of the tangent to a curve at some point can be used for the slope of the curve at that point. Compare the slopes of the two tangents in figure 2. You can see that the slope of the tangent to a point becomes greater as x increases. (Note: In mathematics a straight line can also be referred to as a curve. For example, you may be asked to draw the curve of 2 2y x , which is a straight line.)



Figure 2 Now we need to find a way to determine the slope of a tangent to a curve at any point. Consider figure 3. Points A and B are different points on the curve. The line that passes through points A and B is called the secant line.

Figure 3



In figure 4 you can see that if B approaches A, the secant line (dotted line) becomes the tangent (solid line) to the curve at point A.

Figure 4 Therefore the limiting value of the slope of the secant line will be equal to the slope of the tangent line. To answer our question on how to find the slope of the tangent line, we first need to develop the concept of limits.

3. WHAT IS MEANT BY A LIMIT?

Let’s first get a “feeling” for limits by discussing an example.

Consider the function 2 1f x x when x is near to 2 but not equal to 2.

If we tabulate values of x which approach 2 with the corresponding values of the function f(x), we observe that the closer x comes to the value 2, the closer f(x) comes to the value 3. Table 1

x < 2 x > 2 f(1,7) = 2,4 f(1,8) = 2,6 f(1,9) = 2,8

f(1,99) = 2,98 f(1,999) = 2,998

f(2,3) = 3,6 f(2,2) = 3,4 f(2,1) = 3,2

f(2,01) = 3,02 f(2,001) = 3,002



Figure 5 Both table 1 and figure 5 illustrate that the nearer we take the value of x to 2, the nearer the value of f (x) lies to 3. Note also that it does not matter whether x approaches 2 from the left (x < 2) or from the right (x > 2). We can summarise our observations by saying that 3 is the limit of f(x) when x approaches 2. We write this as

2lim 2 1 3x

x

.

In general we write lim

x af x L

to say that the limit of f (x), as x approaches a, is L.

Finding limits by using graphs or tables is a tedious process. In many cases it would be a lot easier to use our knowledge of algebra to determine a limit.

4. TECHNIQUES FOR FINDING LIMITS

Rules 1 and 2 for limits lim , where and are real numbers

lim , where is a real number

x a

x a

C C a C

x a a

Examples:

2 3lim 7 7 and lim 3x x

x

In the following rules we will assume that

1 2 1 2lim and lim where and are real numbers.x a x a

f x L g x L L L



Rule 3: Limit of a sum or difference

1 2

lim lim limx a x a x a

f x g x f x g x

L L

Thus to find the sum (or difference) of two functions, you can find the sum (or difference) of their limits. Example:

3 3 3

lim 2 lim lim 2 rule 3

3 2 rule 1 and 2

5

x x xx x

Rule 4: Limit of a product

1 2

lim . lim .lim

.

x a x a x af x g x f x g x

L L

This rule states that the limit of the product of two functions is the product of their limits.

Example:

1 1 1

lim3 lim3 lim rule 4

3 1 rule 1 and 2

3

x x xx x

Rule 5: Limit of a quotient

12

2

limlim ,if 0

limx a

x ax a

f xf x LL

g x g x L

If the limit of the denominator is not 0, then the limit of the quotient of two functions is the quotient of their limits.

Example:

2

22

lim 2 42 4lim rule 5

1 lim 1

2(2) 4rule 1 to 4

(2) 1

x

xx

xx

x x

Rule 6: Limit of nf x or n f x

If n is a positive integer



1

1

lim lim

lim lim

nn n

x a x a

nn nx a x a

f x f x L

f x f x L

Examples: 3

33

5 5

3 33

2 2

lim lim 5 and

lim 4 lim 4 8 2

x x

x x

x x

x x

Example 1

3 3

3 3 3

3

3 2 3 2

3 3 3 3

3 2

22

2

22

2

2 2

2 2

2

a) lim 2 5 lim 2 lim5

2 3 5 3

39

b) lim 4 6 lim lim 4 lim 6

3 4 3 6

3

lim 11c) lim

4 1 lim 4 1

lim lim1

lim 4 lim1

(2) 1

(4)2 1

5

7

x x x

x x x x

x

xx

x x

x x

x x x x

x x x x

xx

x x

x

x

ACTIVITY 1 Determine

a) 2

2limy

y y

b) 3

lim 3 4x

x x

c) 2

5lim3k

k

d) 2

41

2 5lim

3x

x x

x




4.1 When the limit of the denominator of a quotient is 0

To find the limit of 2

0

2 4limx

x

x

we find that we cannot use rule 5. Checking the

denominator we see that 0

lim 0x

x

. But this does not mean that the limit does not

exist. If we simplify the fraction, we find that

2 2

2

2 4 4 4 4

4

4

4

x x x

x x

x x

x

x x

x

x

We are interested in the limits as x approaches 0 and not when x has the value 0. We may thus divide by x, that is cancel x in the above manipulation.

Now

2

0 0

2 4lim lim 4 4x x

xx

x

Thus whenever substitution results in 00 , we must do more work to determine whether

the limit exists.

Example 2

2

2 2

2 24a) lim lim

2 2x x

x xx

x x

Note that the denominator becomes 0 and starts to simplify the fraction.

The factor 2x can be cancelled.

This factor is sometimes called the vanishing factor.

Thus

2

2 2

2

2 24lim lim

2 2

lim 2

2 2

4

x x

x

x xx

x x

x

23 3

3

3 3b) lim lim

3 39

1lim

3

1

6

x x

x

x x

x xx

x



1 1

1

1 1c) lim lim Treating 1 as the difference between squares

1 1 1

1lim

1

1

1 1

1

2

x x

x

x xx

x x x

x


a) 2

22

3 10lim

5 14x

x x

x x

b) 2

22

8 12lim

3 10x

x x

x x

c) 2

30

3 2lim

1x

x x

x


4.2 Limits at infinity

Consider 1

lim , 0nx

nx

. Now 1 1 1

lim lim limn n

nx x xx xx

Remember that 1

yx

is the standard form of a rectangular hyperbola. Using the

graph of the hyperbola we find that 1

lim 0x x

.

1 1 1Thus lim lim lim 0 0

n nn

nx x xx xx

.

Notice from this example that as 2 3 4

1 1 1, , , ,...x

x x x all have limits of 0.

Using this with rule 3, we see that 1lim lim lim .0 0

n

nx x x

cc c

xx

.

Thus we have the following which we will use to find limits at infinity: Limits at infinity

If c is a constant then lim 0nx

c

x



Example 3

a) 2lim 3 2x

x x

Take out the highest power of x as a common factor.

22

2

3 2lim 1

1 0 0

xx

x x

b) 2

3

3 2lim

1x

x x

x

We will first divide both the numerator and denominator by the largest power

of x in the denominator, in this case 3x . This will make each term into a constant or a term with a variable in the denominator and allow us to use the properties for limits at infinity.

2 3

3

2

2 3 3 3

3 3

3 3

31 2

1

3 23 2

lim lim1 1

lim1

x x

x x x

xx

x xx x x x x

x x

x x

Now, according to the properties for limits at infinity, all the terms with x in the denominator have a limit of 0. So we have

2

3

3 2 0 0 0lim

1 01

0

1

0

x

x x

x

Hint: When you want to find a limit at infinity, divide both the numerator and denominator by the largest power of the variable in the denominator.

ACTIVITY 3 Evaluate

a) 3 2

3

3 4 1lim

2 2 1x

x x x

x x

b) 2 3 4lim 4 5 4x

x x x




5. ONE-SIDED LIMITS AND CURVES

Figure 6 shows the graph of a function f.

Figure 6 Notice that f (x) is not defined when x = 0. As x approaches 0 from the right, f (x) approaches 1. We write this as

0lim 1x

f x

.

On the other hand, as x approaches 0 from the left, f (x) approaches –1 and we write

0

lim 1x

f x

.

Limits like these are called one-sided limits. From the last section we know that the limit of a function as x a is independent of the way x approaches a. Thus the limit will exist if both one-sided limits exist and are equal. We therefore conclude that

0

limx

f x

does not exist.

As a second example of a one-sided limit, consider 3 as approaches 3

(see figure below). Since is defined only when 3, we speak of the limit as

approaches 3 from the right. From the dia

f x x x

f x x

3gram it is clear that lim 3 0.

xx

Figure 7



A third example: Refer to figure 8.

2

1Now let's look at near 0. Figure 4 below shows the graph of the function.

Notice that as 0, both from the left and from the right, ( ) increases without bound.

Hence no limit exists at

y f x xx

x f x

20

0. We say that as 0, ( ) becomes positively infinite

1and symbolically we write lim

x

x f x

x

Figure 8 A fourth example as shown in figure 9: the graph of the hyperbola.

1Consider the graph of for 0.

1As approaches 0 from the right, becomes positively infinite.

y f x xx

xx

Figure 9



0 0

1As approached 0 from the left, becomes negatively infinite.

1 1Symbolically we write lim and lim .

x x

xx

x x

Either one of these facts implies that 0

1limx x

does not exist.

6. RESPONSES TO ACTIVITIES

6.1 Activity 1

a) 2 2

2 2 2

2

lim lim lim

2 2

6

y y yy y y y

b)

3 3 3

3 3 3 3

lim 3 4 lim 3 .lim 4

lim lim3 . lim lim 4

3 3 3 4

6

x x x

x x x x

x x x x

x x

c)

2 2

5 5lim3 3lim

3 25

75

k kk k

d)

22

14 41

1

lim 2 52 5lim

3 lim 3

2

2

1

x

xx

x xx x

x x



6.2 Activity 2

2

22 2

2

3 5 23 10a) lim lim

7 25 14

3 5lim

7

3 2 5

2 7

11

9

x x

x

x xx x

x xx x

x

x

2

22 2

2

2 68 12b) lim lim

2 3 53 10

6lim

3 5

2 6

3(2) 5

4

11

x x

x

x xx x

x xx x

x

x

22

3 30

0 3 0 23 2c) lim

1 0 1

2

1

2

x

x x

x

Note in this case we can use rule 5, as the denominator does not become 0.



6.3 Activity 3

a)

2 3

2 3

3 2

3

4 1 1

2 1

3 4 1lim

2 2 1

3lim

2

3

2

x

x x x

xx x

x x x

x x

b) 2 3 4lim 4 5 4x

x x x

2 3 44

4 4 4 4

44 2

4

4

4 5 4lim

4 5 1lim 4

0 0 0 4

4

x

x

x x xx

x x x x

xxx x

This is the end of learning unit 2 on limits and so you should be able to explain what is meant by a limit find the value of a limit give examples of one-sided limits We now move on to learning unit 3 on the derivative.


Differentiation The derivative

CONTENTS PAGE

1. THE TANGENT QUESTION AND THE DERIVATIVE .................................. 27 2. DIFFERENTIATION FROM FIRST PRINCIPLES ........................................... 28 3. RESPONSE TO ACTIVITY ................................................................................ 31

MODULE 6

LEARNING UNIT 3

OUTCOMES

At the end of this learning unit, you should be able to relate the rate of change of a function to the gradient of the tangent at a point differentiate simple expressions from first principles recognise the different notations used to denote derivatives

Module 6 Learning unit 3 DIFFERENTIATION: The derivative


1. THE TANGENT QUESTION AND THE DERIVATIVE

We return to the discussion of the tangent question mentioned in learning unit 2. We want to find an expression for the slope of the curve, or the slope of the tangent at a point on the curve. Consider figure 1.

Figure 1

We can write the slope (gradient) of the secant line through A and B as

change in

change in AB

f x h f xy ym

x x h

If we move B closer to A, h becomes smaller and is getting closer to 0 and the secant line becomes the tangent. The slope of the tangent line will thus be the limit

0

limh

f x h f x

h

This formula gives us the slope of the curve at point A. This special limit is called the derivative of a function. The mathematical process of finding the expression for the gradient of a curve at any point is called differentiation.

Notation

The symbol d

dxconsidered on its own is called the differentiating operator, and

indicates that any function written after it is to be differentiated with respect to x. We



always differentiate with respect to the independent variable, that is

dependent variable

independent variable

d

d.

Note: dy y

dx x

y

x

is the average rate of change over an interval (slope of the secant) while

dy

dx is the limiting value equal to the instantaneous rate of change at a point

(slope of the tangent). The following symbols are often used to indicate derivatives instead of

0

limh

f x h f x

h

:

i we say by ii ' we say prime

iii iv '

v vix

dydy dx f x f x

dxd

f x ydxD y Df x

dy

dx has various interpretations:

1. the change in y due to the change in x 2. the differentiation of y with respect to x 3. the first differential 4. the first derivative 5. the slope or gradient 6. tan where is the angle of inclination of the tangent to the horizontal; see

figure 2

Figure 2

2. DIFFERENTIATION FROM FIRST PRINCIPLES

Differentiation from first principles means to calculate

0limh

f x h f x

h

. We will

take four steps to calculate this special limit:



1. Find f x h

2. Find f x h f x

3. Divide by h 4. Let 0h to find the limit

Example 1 a) Find the expression for the derivative of y = x2. Solution We take the steps as set out above:

2

2

2 2

2 2 2

2

Step 1: Thus

2

Step 2: 2

2

f x x

f x h x h

x xh h

f x h f x x xh h x

xh h

22Step 3:

2

2

f x h f x xh h

h hh x h

hx h

0 0

0

Step 4: lim lim 2

2

Thus the derivative = = lim 2

h h

h

f x h f xx h

hx

f x h f xdyx

dx h

This example can be taken a step further: b) Determine the derivative of y = x2 at x = 3.

Solution Substitute the given value into the expression obtained in (a):

3

2(3) 6x

dy

dx

.

Where 3x

dy

dx

is a shorthand notation for the derivative at x = 3.



We can say that the slope or gradient of the curve at x = 3 is 6.

Example 2

12If 2 find ' from first principles.f x x f x

Solution

1

2

12

3221 11 1

2 212 22 2

1 1 322 2 2

Step 1: 2

The factor may be expanded by using the binomial theorem

2 .....

1 12 ....higher order terms in

2 2

Step

x h

f x h x h

x h

x x h

x x h x h h

1 1 3 122 2 2 2

1 322 2

1 3

2 2

1 1 2: 2 ..... 2

2 2

1 1....higher order terms in

2 2


2 2Step 3:

f x h f x x x h x h x

x h x h h

h x x h hf x h f x

h h

x

1 3

2 2

1 3

2 2

0 0

1

2


2 2

1 1Step 4: lim lim .. ....higher order terms in

2 2

All the terms containing an become 0.

h h

x h h

f x h f xx x h h

h

h

x



ACTIVITY 1 Determine the derivatives of the following from first principles:

a) 23 at 2y x x

b) 32 2f x x

c) 1f x x

x

d) 8y x (Hint: Use the binomial theorem.) Remember to check the response on page 31.


Activity 1

a) 2Put 3y f x x

2

2 2

2 2

2 2 2

2

2

0 0

2

Step 1: 3

3 2

3 6 3

Step 2: 3 6 3 3

6 3

6 3Step 3:

6 3

6 3

Step 4: lim lim 6

6

6 2 12

h h

x

f x h x h

x xh h

x xh h

f x h f x x xh h x

xh h

f x h f x xh h

h hh x h

hx h

f x h f xx h

hx

dy

dx



b) 33 2

2 2

8 24 24 8

f x x

x x x

3

3 2

3 2 2 3 2 2

3 2 2 3 2 2

3 2

2 2 3 2

Step 1: 2 2

8 24 24 8

8 24 24 8 24 48 24 24 24 8

Step 2: 8 24 24 8 24 48 24 24 24 8

8 24 24 8

24 24 8 48 24 24

Step

f x h x h

x h x h x h

x x h xh h x xh h x h

f x h f x x x h xh h x xh h x h

x x x

x h xh h xh h h

2 2

2 2

2 2

0 0

2

2

22

24 24 8 48 24 243:

24 24 8 48 24 24

Step 4: lim lim 24 24 8 48 24 24

24 48 24

6 2 2

Thus ' 24 48 24 6 2 2

h h

h x xh h x hf x h f x

h h

x xh h x h

f x h f xx xh h x h

h

x x

x

f x x x x

c)

2 2

1

1Step 1:

1 1Step 2:

1 1

f x xx

f x h x hx h

f x h f x x h xx h x

hx h x

xh x h x x h

x x h

x h xh h

x x h



2 2

2

2

Step 3:

1 1

1

f x h f x x h xh hh

h x x h

h x xh

x x h h

x xh

x x h

2

0 0

2

2

2

2

1Step 4: lim lim

1

11

1' 1

h h

f x h f x x xh

h x x h

x

x

x

f xx

d) 8Put y f x x

8

8 7 6 2

8 7 6 2 8

7 6 2

7 6

7 6

Step 1:

8 28 .....

Step 2: 8 28 .....

8 28 ....higher order terms in

8 28 ....higher order terms in Step 3:

8 28 ....higher orde

f x h x h

x x h x h

f x h f x x x h x h x

x h x h h

h x x h hf x h f x

h h

x x h

7 6

0 0

7

r terms in

Step 4: lim lim 8 28 ....higher order terms in

8

h h

h

f x h f xx x h h

h

x

7Thus 8dy

xdx

.

You have completed this learning unit and so you should be able to relate the rate of change of a function to the gradient of the tangent at a point differentiate simple expressions from first principles recognise the different notations used to denote derivatives

In learning unit 4 you will learn about standard forms.


DIFFERENTIATION Standard forms

CONTENTS PAGE

1. STANDARD FORMS .......................................................................................... 35 2. RESPONSES TO ACTIVITIES ........................................................................... 39 2.1 Activity 1 .............................................................................................................. 39 2.2 Activity 2 .............................................................................................................. 39

MODULE 6

LEARNING UNIT 4

OUTCOMES

At the end of this learning unit, you should be able to differentiate a function in the standard form without referring to your notes.

Module 6 Learning unit 4 DIFFERENTIATION: Standard forms


1. STANDARD FORMS

The process of finding the derivative is called differentiation, and deriving it via the limit is called differentiation from first principles. It is quite an involved process and to avoid having to evaluate limits every time we differentiate, we use established derivatives called standard forms The standard derivatives summarised below may be proved theoretically and are true for all real values of x. Notice that the list includes derivatives of the trigonometric functions. In the forms in which the derivatives are given, it is essential that the angles be measured in radians. You must memorise this list. or y f x or '

dyf x

dx

1 constant, k 0

2 x 1

3 nx 1nnx is a constantn

4 nax 1nanx and are constantsa n

5 xe xe

6 kxe kxke is a constantk

7 xa nxa a is a constanta

8 n x 1

x

9 loga x 1

nx a

is a constanta

10 sin kx cosk kx is a constantk

11 cos kx sink kx is a constantk

12 tan kx 2seck kx is a constantk

13 cot kx 2coseck kx is a constantk

14 sec kx sec tank kx kx is a constantk

15 coseckx cosec .cotk kx kx is a constantk



Example 1 Differentiate a) y = 6

b) y = 6x Solution a) 6 is a constant. Use standard form 1.

0dy

dx .

b) Use standard form 4 with a = 6 and n = 1.

1 1 06 1 6 6dy

x xdx

Example 2

Find the derivatives of a) 312y x

b) 3

12y

x

c) 3y x Solution

Use standard form 4: If ny ax then 1ndyanx

dx .

a) a = 12 and n = 3 thus 3 1 212 3 36dy

x xdx

.

b) 3

12y

x is rewritten in the standard form nax as 312y x . So a = 12 and n =

3, thus 3 1 44

3612 3 36

dyx x

dx x

c) 3y x is rewritten in the standard form nax as 1

23y x . So a = 3 and n =

1

2, thus

1 11

2 21

2

1 3 3 33

2 2 22

dyx x

dx xx

.

Example 3

Find 'f x if a) 2xf x

b) nf x x

c) 3xf x e

d) 2logf x x

Solution

a) Refer to standard form 7 with a = 2. Thus ' 2 n 2xf x .

b) Refer to standard form 8. Thus 1'

xf x .

c) Refer to standard form 6. Thus 3' 3 xf x e .



d) Refer to standard form 9. Thus 1'

n 2f x

x

.

Example 4

Determine dy

dx if: a) siny x

b) sec3y x

c) cos4

xy

Solution

a) Refer to standard form 10. 1k Thus cosdy

xdx

b) Refer to standard form 14. 3k Thus 3sec3 tan 3dy

x xdx

c) Refer to standard form 11. 14k Thus

1sin

4 4

dy x

dx

ACTIVITY 1 Determine the derivatives to x of the following:

5

3

10

5 6

3

3104

a) b) sin 3

c) cot 2 d)

e) log f) n 3

g) h) 5

i) tan 3 j) 10

k) cosec l) log

m) n) 6

x

x

x x

x x

x x

e x

x

x x

x x

4

o) p) sec4 2

x x

Remember to check the response on page 39 The table is written with independent variable x. However, it can still be used when other independent variables are involved, as shown in the next example.

Example 5

Find the derivatives of a) sin5y t t

b) n3y t t

c) 5f s s

Solution



1 4

5 5 54 5 45

a) ' 5cos5

(3) 1b) '

3

1 1 1c) Thus '

5 55

y t t

y tt t

f s s s f s sss

ACTIVITY 2 1. Determine the derivatives of the following:

3

3

a) cos ec 5

b) 6

c)

d) n

e) log

y t t

g t t

f s s

u w w

V h h

2. Differentiate the following:

32

7

2

2

a) n 2

b)

c)

1

1e)

f)

g)

d)

y

f x x

r h

t

wu

f t x

f x x

v

Remember to check the responses on pages 39 and 40.




2.1 Activity 1

4

2

2

32 3 23

5

5

2

a) 5

b) 3 cos 3

c) 2cosec 2

1 1 1d)

3 33

1e)

n10

1f)

3

g) 5

h) 30

i) 3sec 3

j) 10 n10

k) cos ec .cot

1l)

n 3

x

x

x

x

x

xx

x

x

x

e

x

x

x x

x

194

3 3

3m) n) 60

41 1

o) 4 p) sec tan4 2 2 2

x x

x xx x

2.2 Activity 2

1.

2

1

2

a) ' 5cos ec 5 .cot 5

b) ' 18

1 1c) '

2 21

d) '

1e) '

n 3

y t t t

g t t

f s ss

u ww

V hh



2.

a) 0dy

dx , because n 2 is a constant. You can find the value with your

calculator.

b) 6' 7f x x

c) 32

1

213 3

2 2

drh h

dh . In this case r is the dependent variable and h the

independent variable.

d) 1 1 3

12 2 21 1

Thus 2 2

dvt t

dtv t

.

e) 2 32 3

1 2. Thus 2

dww u u

duu u

.

f) f t x . Be careful here - the independent variable is given as t; thus x must

be regarded as a constant in this case.

' 0f t

g) 2 1' 2f x x

Now that you have reached the end of learning unit 4, you should be able to

differentiate a function in the standard form without referring to your notes. The rules of differentiation will be covered in learning unit 5 next.


DIFFERENTIATION Rules of differentiation I

CONTENTS PAGE

1. DERIVATIVES OF COMBINED EXPRESSIONS ............................................ 42 1.1 Constant times a function ...................................................................................... 42 1.2 Sums and differences ............................................................................................ 42 1.3 Product rule ........................................................................................................... 45 1.4 Quotient rule ......................................................................................................... 46 2. RESPONSES TO ACTIVITIES ........................................................................... 47 2.1 Activity 1 .............................................................................................................. 47 2.2 Activity 2 .............................................................................................................. 47 2.3 Activity 3 .............................................................................................................. 48 2.4 Activity 4 .............................................................................................................. 49 2.5 Activity 5 .............................................................................................................. 49

MODULE 6

LEARNING UNIT 5

OUTCOMES

At the end of this learning unit, you should be able to use the following rules of

differentiation to differentiate combined expressions: constant times a function rule for sums and differences product rule quotient rule

Module 6 Learning unit 5 DIFFERENTIATION: Rules of differentiation I


1. DERIVATIVES OF COMBINED EXPRESSIONS

The rules of differentiation exist for finding the derivatives of functions that have been combined under various operations. The mathematics involved in proving these rules can become complicated. For our purposes it will suffice to present the rules without proof. Although there are many functions for which the derivative does not exist, our concern will be with functions that are differentiable.

1.1 Constant times a function

When the derivative of a function is already known, it is a simple matter to find the derivative of a constant multiple of that function. This involves using the rule

If then ' ' where is a constant.F x k f x F x k f x k

That is, the derivative of k times a function is simply k times the derivative of the function.

Example 1

3 2 2

5 6 6

5 5 31

2 2 2

a) If 9 then ' 9 3 27

b) If 7 then ' 7 5 35

2 2 5 5c) If then

3 3 2 3

f x x f x x x

g x x g x x x

dyy x x x

dx

ACTIVITY 1 Find the derivatives of

2

4

3

a) 3

b) 12

c) 2

f x x

y x

V y


1.2 Sums and differences

If then ' ' '

If then ' ' '

F x f x g x F x f x g x

F x f x g x F x f x g x

According to these rules, the derivative of a sum (or difference) is the sum (or difference) of the individual derivatives. To use these rules, we differentiate each term.



Example 2 Find the derivatives of

2

5 3

2

a) sin

b) 7 4 2 8

c) nx

F x x x

f x x x x

y e x x

Solution Differentiate the given functions term-by-term using the correct standard forms:

4 2

12 22

a) ' 2 cos

b) 35 12 2

1 1 1 1c) 2 2

2 2x x

F x x x

f x x x

y e x ex x x


2

2

2

a) if 3 5 6

4b) ' if 2

3c)

2

dyy x x

dx

f x f x xx

d xx

dx x

Remember to check the response on page 47. In some cases a quotient or product can be simplified to the sum or difference of functions:

Example 3 Find the first derivative of

2 5

3

2

3 4a)

b) 3 2

t tS

t

y x

Solution

2 5 2 5

3 3 3

1 2

3 4 3 4a)

3 4

t t t tS

t t t

t t



1 1 2 1

2

2

3

2

3

2

3 1 4 2

3 8

3 8

1

3 8

8 3

dst t

dt

t t

t

t

t

t

t

t

If possible, we simplify answers to the same format as the original question.

2

2

2 1 1 1

0

b) 3 2

9 12 4

9 2 12 1 0

18 12 (Remember 1)

6 3 2

y x

x x

dyx x

dx

x x

x

ACTIVITY 3 1. Differentiate with respect to the independent variable:

3 24 3

3 14 4

2 72

2

a) 4 3 2 b)2 7

2 3c) d) 2 3

2e) 2 4 f)

2

d d z zx x

dx dz

y f t t tx x

xy x x y

x

2

2. Find in the following cases:

a)

b)

dy

dx

a bx cxy

xa

y axax




1.3 Product rule

Note that the derivative of a product does not behave as nicely as other products you have encountered thus far. The derivative of a product is not the product of the derivatives. The product rule is given by the following formula:

If then ' ' 'F x f x g x F x f x g x f x g x

In words, we have the derivative of product equals the first function times the derivative of the second function plus the second function times the derivative of the first function.

Example 4

Determine if 3 1 2 1dy

y x xdx

Solution

Let 3 1 and 2 1

' 3 and ' 2

' '

3 1 2 3 2 1

6 2 6 3

12 5

f x x g x x

f x g x

dyf x g x f x g x

dxx x

x x

x

Example 5

If 2 sin find .dy

y x xdx

Solution

2

2

2

Let and sin

cos 2 sin

cos 2 sin

f x x g x x

dyx x x x

dx

x x x x

ACTIVITY 4 Determine the derivatives of

2

2

a) 3 5 2

b) cos

c) x

y x x x

f t t t

y xe




1.4 Quotient rule

The quotient rule is given by the following formula:

2

' 'If then '

f x g x f x f x g xF x F x

g x g x

Consider the following examples:

Example 6

sinIf find .

x dyy x

x dx

Solution

2

2

2

Let sin and

Then ' cos and ' 1

' 'Thus

cos sin (1)

cos sin

f x x g x x

f x x g x

g x f x f x g xdy

dx g x

x x x

x

x x x

x

Example 7

1Find ' if

1

xy y

x

Solution

1 1

2 2

1 1

2 2

Let 1 +1 and -1

1 1Then ' and '

2 2

f x x x g x x

f x x g x x

1 1 1 12 2 2 2

12

2

1 12 2

2

' 'Thus '

1 1

1

g x f x f x g xy

g x

x x x x

x

1 1 1 1 1 12 2 2 2 2 2

12

1 1 1 12 2 2 2

21

x x x x

x



1 12 2

12

12

12

1 1 1 12 2 2 2

2

2

2

1

1

1

1

x x

x

x

x

x x

ACTIVITY 5 Determine the derivatives of

2

2

2a)

2 3

b)2

tanc)

2

d) ( )n

t

xy

x

ef t

tx

yx

xf x

x



2.1 Activity 1

3

2

a) ' 3 2

b) 12 4

c) 2 3

f x x

y x

V y

2.2 Activity 2

1 1

2 2

2

2 1 1 1

a) 3 5 6

3 2 5 1 0

6 5

4b) 2

2 4

y x x

dyx x

dxx

f x xx

x x



1 12 2

312 2

312 2

32

1 1

3

1 1' 2 4

2 2

2

1 2

2

2

2

f x x x

x x

x xx

xx

xx

x x

2 22

2

2 12 1 1 1

3

3

3c) 3

2 2

23 2 1

2

6 1

61

d x d xx x x

dx dxx

xx x

x x

xx

2.3 Activity 3

1134

514 4

6

2 3

2

1. a) 3 4

b)

2 6c)

3d) ' 2

23

e)2

1 1f)

4

2. a)

b)2 2

x

z z

dy

dx x x

f t t t

dyx x

dxdy

dx x x xdy a

cdx xdy a a

dx ax x ax



2.4 Activity 4

a) 2

2 2

2

3 5 . 2 2 3. 2

6 4 10 3 6

9 2 10

dyx x x x

dx

x x x x

x x

1

21b) ' sin .cos

2cos

.sin2

f t t t t t

tt t

t

2 2

2 2

2

c) (2 ) (1)( )

2

2 1

x x

x x

x

dyx e e

dx

xe e

e x

2.5 Activity 5

2

2

2

2

2 2

2

2

2

2

2

2a)

2 3

Let ( ) 2 and ( ) 2 3

' 2 ' 3

2 3 2 2 3

2 3

4 6 6 3

2 3

3 4 6

2 3

3 4 6

2 3

xy

x

f x x g x x

f x x g x

x x xdy

dx x

x x x

x

x x

x

x x

x



2

2

22

2

22

b)2

2 2'

2

2 2

2

t

t t

t

ef t

t

t e e tf t

t

e t t

t

2

2

2

2

2

2

tanc)

2

2 sec tan 2

2

2 sec tan

4

sec tan

2

xy

x

x x xdy

dx x

x x x

x

x x x

x

2

2

d) ( )n

1n 1

'n

n 1

n

xf x

x

x xxf x

x

x

x

Learning unit 5 is now complete and you should be able to differentiate combined

expressions using the following rules of differentiation: constant times a function rule for sums and differences product rule quotient rule We examine the next part of the rules of differentiation in learning unit 6.


DIFFERENTIATION Rules of

differentiation II

CONTENTS PAGE

1. DERIVATIVES OF COMPOSITE FUNCTIONS .............................................. 52 1.1 Function-of-a-function or chain rule ..................................................................... 52 1.2 General standard forms ......................................................................................... 61 2. RESPONSES TO ACTIVITIES ........................................................................... 65 2.1 Activity 1 .............................................................................................................. 65 2.2 Activity 2 .............................................................................................................. 65 2.3 Activity 3 .............................................................................................................. 66 2.4 Activity 4 .............................................................................................................. 66 2.5 Activity 5 .............................................................................................................. 68

MODULE 6

LEARNING UNIT 6

OUTCOMES

At the end of this learning unit, you should be able to differentiate composite functions using the chain rule together with the rules for combined expressions: constant times a function rule for sums and differences product rule quotient rule

Module 6 Learning unit 6 DIFFERENTIATION: Rules of differentiation II


1. DERIVATIVES OF COMPOSITE FUNCTIONS

Suppose we are given a function y(x), where the variable x is itself a function of another variable, t say. We say that y is a function of a function.

Suppose 3 and sin ,y x x x t t we can write 3siny t . There are two

functions working on the variable t. When we differentiate, both functions must be considered.

1.1 Function-of-a-function or chain rule

The chain rule states:

If ( ) then ' ' ( ) . '( ).F x f g x F x f g x g x

It is often easier to make a substitution before differentiating and using the d

dx

notation. The chain rule is then stated as follows:

Given a function ( ) where ( ) then y y x x x t

dy dy dx

dt dx dt

Note: We have stated the rule for two functions but any number of functions can be involved. We will first look at examples where powers of a function are involved.

Example 1

Find dy

dx if 43 1y x .

Solution

3

33 2

32 3

Let ( ) 1

Then from the chain rule

' ( ) . '( )

4 1 3

This answer can be simplified.

=12 1

g x x

dyf g x g x

dx

x x

x x



Example 2

If 3( ) siny t t find

dy

dt.

Solution

Make the substitution sinx t t

Thus 3y x . Now differentiate each of the functions:

23 and cos dy dx

x tdx dt

Then using the chain rule we can write:

23 cos

dy dy dx

dt dx dt

x t

Remove the substitution by substituting sinx t .

2

2

3 sin cos

3sin cos

t t

t t

Example 3

Differentiate 21y x . Solution

We note that 21y x is a function, namely the square root of another function, 21 x .

We can make a substitution:

1

2 2Let 1 then u x y u u

To use the chain rule we require and du dy

dx du.

1

21 and = 2

2

dy duu x

du dx

Thus

1

2

1

2

12

2

dy dy du

dx du dx

u x

xu

Substitute 21u x



12 2

12 2

2

1

1

1

x x

x

x

x

x

In the following examples we will not write out the substitution. If you battle to follow this approach, write out the substitutions and compare your answer to the answer in the notes.

Example 4

Find the derivative of 3

2 24 1y x .

Solution

312 22

12 2

12 2

4 3. 1 1

1 2

6 1 2

12 1

dy dx x

dx dx

x x

x x

Example 5 2

1Find if

dyy

d

.

Solution

221

2 11 1

11 2

2

3

3

1

2

2 1 1

1 12 1

1 1 12

12

S

dy d

d d



ACTIVITY 1 Differentiate with respect to the independent variable:

83 2

2

32

3

3 2

3

2

1a) b)

1

c) 1 d) 2 3

1e) 4 9 f)

3

g) and are constants

s f y y yt

s x f t t

y x yx

by a a b

x

Remember to check the response on page 65. In the following example we apply the product rule as well as the function-of-a-function rule:

Example 6

2Find if 3 2 5 4.ds

s t tdt

Solution

12

2

2

3 2 5 4

3 2 5 4

s t t

t t

We will use a substitution for the product rule but not for the function-of-a-function part.

12

12

12

12

Function of a function

2

1

Let 3 2 and 5 4

16 and 5 4 5 4

2

15 4 5

2

55 4

2

u t v t

du dv dt t t

dt dt dt

t

t

Writing the product rule in d

dt notation we have

1 1

2 22

. .

53 2 . 5 4 5 4 . 6

2

ds dv duu v

dt dt dt

t t t t



There are various methods to simplify. We will illustrate two methods: Method 1

1 1

2 22 53 2 . 5 4 5 4 . 6

2

dst t t t

dt

Take out the highest common factor (HCF) in both terms:

1 1 12 2 2

12

2

HCF

2 2

2

15 4 5 3 2 12 5 4

2

15 4 15 10 60 48

2

75 48 10

2. 5 4

t t t t

t t t t

t t

t

Method 2

1 1

2 22 53 2 5 4 5 4 6

2

dst t t t

dt

Write the first term as a fraction. Add terms using rules for handling fractions.

12

12

1 12 2

12

12

2

2

2 2

2

53 2 5 4 6

2 5 4

5 3 2 12 5 4

2 5 4

15 10 60 48

2 5 4

75 48 10

2. 5 4

t t tt

t t t

t

t t t

t

t t

t

ACTIVITY 2 Differentiate with respect to x and simplify:

3

4

5 2

2

2 2 4 3

a) 1 2 b)

c) 2 1 1 d) 3 4

e) f) 2 7

y x x r x ax b

s x x r

x x a y x x




In the following examples we will use the quotient rule and the chain rule.

Example 7

3

4If find

1

x dyy

dxx

and simplify.

Solution We recognise the given function as a quotient and proceed to use the quotient rule. [Note: we could handle this question as a product rule by rewriting y as

43 1y x x .]


4 32 3

24

H.C.F.

32

8

32

8

2

5

1 3 . 4 1 1

1

1 3 1 4

1

1 3 3 4

1

3

1

x x x xdy

dx x

x x x x

x

x x x x

x

x x

x

Example 8

2

2

3 2 1Determine ' if

1

xf x f x

x

and simplify the answer.

Solution

12

1 12 2

12

2

2

2

2


2 2 212

22

3 2 1

1

3 2 1

1

1 .3 4 3 2 1 . 1 2'

1

xf x

x

x

x

x x x x xf x

x



There are many different ways to simplify. All of them yield the same answer. We will use the following method:

Multiply the top and bottom by the lowest common multiple (LCM) of all the

denominators in the top and bottom. LCM in this example is 1

2 21 x .

12 22

12 2

1 1 12 2 2 22 2 2

12 2

12 2 2

12

32

12 1 3 2 1

11

2

12 1 1 3 2 1 1

1 1 11

1 1

1 1

12 2

12

2 2

2

2

32

'1

12 1 3 2 1

1

3 4 4 2 1

1

3 3 2

1

x x x x

x

x x x x x x

x

x x

f xx

x x x x

x

x x x

x

x x

x

Example 9

Find the derivative of 21 x

yx

using

a) quotient rule b) product rule Solution

a)

122 21

2

2

. 1 . 2 1. 1x x x xdy

dx x

2

2

2 2

2

2

12

1

12

2 2

1

1

. 1

x

x

x x

x

x

x

x

x x



b) 1

22

1 211

xy x x

x

1 12 2

12

12

1 12 2

12

12


1 2 2 2

2

22

2 2

2 2

2 2

2 2

2

2 2

11 2 1

2

11

1

1

1

1

1

2 1

1

dyx x x x x

dx

x

xx

x x

x x

x x

x x

x

x x

ACTIVITY 3 1. Differentiate with respect to the independent variable and simplify:

2

2

3

5

43

3

2 1 3a) b)

2 13 1

2 2 3c) d)

5

2 Differentiate by using (i) the quotient rule and

(ii) the product rule

a)1

1b)

2 1

x xy r

sx x

x x xy y

xx

xy

x

xy

x


Example 10

By making a substitution 2 3u x x , use the chain rule to find the derivative of

2n 3y x x .

Solution



2Let 3 then nu x x y u

In this example the chain rule becomes dy dy du

dx du dx .

Now 1

2 3 and du dy

xdx du u

Thus

2

12 3

2 3

3

dy dy du

dx du dx

xu

x

x x

Note that the numerator is the derivative of the denominator. The result of the previous example can be generalised to any function of the form

ny f x . We have

'

If n then .f xdy

y f xdx f x

Example 11

By making a substitution 2 3u x x , use the chain rule to find the derivative of 2 3x xy e .

Solution

2Let 3 then uu x x y e

In this example the chain rule becomes dy dy du

dx du dx .

Now 2 3 and udu dyx e

dx du

Thus

2 3

2 3

2 3

u

x x

dy dy du

dx du dx

e x

x e

The result of the previous example can be generalised to any function of the form f xy e . We have

( )If then '( ) .f x f xdyy e f x e

dx



Example 12

By making a substitution 2 3u x x , use the chain rule to find the derivative of

2sin 3y x x .

Solution

2Let 3 then sinu x x y u

Now 2 3 and cosdu dy

x udx du

Thus 2 3 cosdy dy du

x udx du dx

The result of the previous example can be generalised to any function of the form

siny f x . In fact, using the chain rule we can generalise all the standard forms

given in unit 4 as follows:

1.2 General standard forms

y

dy

dx

1 nf x 1

. 'n

n f x f x

2 (f xe ' f xf x e

3 f xa ' nf xf x a a

4 n f x 'f x

f x

5 loga f x '1

.n

f x

a f x

6 sin f x ' cosf x f x

7 cos f x ' sinf x f x

8 tan f x 2' secf x f x

9 cot f x 2' cosecf x f x

10 sec f x ' sec tanf x f x f x

11 cosec f x ' cosec .cotf x f x f x



ACTIVITY 4 Differentiate the following:

2

32

3 2

a) 1 b) 2 5

c) d) x

y n x y n x

y n n x y e

2

2

2 3

2 5 sin3

5 42

2 2

e) n 3 f) log 2 1

g) 4 h)

i) 10 j) log 3

k) 1 .

x

x x x

x

x

y e e y x

y y e

y y x

y x e

Remember to check the response on page 66. When working with logarithmic or trigonometric functions, it is sometimes better to first simplify the given functions before differentiating.

Example 13 2

2

1Find if .

1

dy xy n

dx x

Solution

22 2

2

2 2

2 2

2 2

3 3

4

4

1 a 1 1 log log log

b1

2 2

1 1

2 1 2 1

1 1

2 2 2 2

1

4

1

xy n n x n x a b

x

dy x x

dx x x

x x x x

x x

x x x x

x

x

x

Example 14

Find the first derivative if tan 1

sec

xy

x

.

Solution

Simplify first:

sin1

tan 1 cos1sec

cos

xx xy

xx



sin cos 1

cos cos

sin cos cos

cos 1

sin cos

x x

x x

x x x

x

x x

The differentiation becomes very easy:

cos sin

cos sin

dyx x

dx

x x

What would happen if we did not simplify first?

2

3 2

2

2

2

3 2

2

sin sin1 1 1cos cos coscos cos1

cos

23

2

tan 1

sec

sec sec tan 1 sec tan

sec

sec tan .sec sec .tan

sec

. .

1 sin sin .cosMultiply top and bottom by cos

cos

sincos sin cos

cos

x xx x xx x

x

xy

x

x x x x xdy

dx x

x x x x x

x

x x xx

x

x x x

x

2 2

2 2

cos 1

cos 1 sin

cos cos sin

cos

cos sin

x x

x x

x x x

x

x x

Problems like examples 13 and 14 do not occur frequently, but they are good questions to ask as they combine different fields. The next activity allows you to revise what you have learnt thus far in differentiation.

ACTIVITY 5 1. Find the 'y and simplify your answers to the same format as the original function, if possible:

64 2

2 2

2 2 2

a) 2 b) 2 4 5

1 4c) d)

2

y x y x x

a xy y

x a xx



2 22

2 2e) f) 3 2a x

y y x xa x

11 3 22 2

2

3 2

g) h) 11 4

i) 4 3 j) 1

wy y x

w

y x x y x x

2. Differentiate with respect to the independent variable and simplify the answer:

2

2

10

3

2 sin

a) log 4 1 b) cos

c) cot d)

3. Find the value of ' for the given value of in radians .

a) tan ;4

b) 5 sin ; 22

c) 10 sin 3 ; 1

x

x

x x

R x p n x

y e s e

f x x

f x n x x

xf x e x

f x e x x

4. Find the first derivative of:

4

2

3 2

2 2

2 4 2

a) sin b) n cos

c) sin 2 d) .sin

1 cose) cos f)

1 cos

g) tan h) tan

i) tan 3 j) sin

y x y x

f x x x x

xy x y

x

y x y x

y x f x x x

5. Differentiate with respect to the independent variable, and simplify

where possible:

2 3

3 2

a) 3sin 4 b) sin 3 3

c) sin d) sin

1e) cos f) tan .sin 2

2

y x y x

y x x

x y x x

Remember to check the response on pages 68 and 69.




2.1 Activity 1

a) 1

22

2

11

1S t

t

12

32

32

12 2

2

2

32

11 1

2

11 2

2

1

1

ds dt t

dt dt

t t

t

t

t

t

Only the answers to the rest of the questions are supplied.

73 2 2

22

23

2

3 2423

1b) ' 8 3 2 c)

2 1

3d) ' 18 2 3 e)

4 9

2 6f) g)

3 3

dsf y y y y y

dx x

dyf t t t

dx x

dy x dy b ba

dx dx x xx

2.2 Activity 2

24

2

3 22 2

42 2 4 3

5 4a) 1 2 1 12 b)

2

2 3 53 6c) d)

2 1 3 4

3 4 62e) f)

4. 2 7

dy dr ax bxx x

dx dx ax b

ds x dr

dx dx

x xd x a dy

dx dxx a x x



2.3 Activity 3

In questions a) to c) it is not necessary to use the chain rule.

2 22

2

3 2

32 34

6 53

5 2 61. a) b)

2 13 1

3 3 45 20c) d)

2 5 2 3

36 152. a) b)

1 2 1

x xdy dr

dx ds sx x

x xdy dy x x

dx dxx x x

x xdy x dy

dx dxx x

2.4 Activity 4

a) 2This function has the form with 1y n f x f x x

2

2Therefore

1

dy x

dx x

b)

3Given 2 5

Note that 3 2 5

y n x

y n x

23

2 5

6

2 5

dy

dx x

x

c) is of the form n where ny n n x y f x f x x

'

Therefore

1

1 1

1

f xdy

dx f x

xn x

x n x

x n x



d) 23 26 xdy

xedx

e) 3 30 3 0 3x xdye e

dx (Since 2 and n3e are constants.)

f) Use the table of standard forms, number 5, with a = 10 and 2 1f x x .

1 2.

log10 2 1

2(Since log 10 = 1)

2 1

dy

dx x

x

g) Use the table of standard forms, number 3, with a = 4 and 22 5f x x x .

2

2

2 5

2 5

4 5 4 . n 4

n 4 4 5 4

x x

x x

dyx

dx

x

It is important to use brackets in your answers to prevent confusion. h) Use the table of standard forms, number 6.

sin 33cos3 xdyx e

dx

i) 25(10 ) 10 n10xdy

xdx

Using a calculator this can be simplified to 2523,03 10 xx .

Note: you usually do not simplify using a calculator when finding derivatives except if the questions asks you to find the value of the derivative.

j) 3

4

1 12.

n 2 3

4

n 2

dy x

dx x

x

k) 1

1 22 22 2 1. 1 .

xxy x e x e



1 12 2

2 2

12

22

12

2 2

12

2

12

2

2 21 12 2

2

2

12

2

2

2

2

2

' 1 . 1 2

1

2 1

1 2

2 1

1 2

2 1

2 1

2 1

x x

xx

x x

x

x

y x e e x x

e x xe

x

e x xe

x

e x x

x

e x x

x

2.5 Activity 5

1.

11 1 22 2

53 2

2

3 23 2 2

2 2

2 2 4 4 2

32

2 2 212

a) ' 4 2 b) ' 24 1 2 4 5

1 2 4c) ' d) '

2 3 4e) ' f) '

3 2

1 1g) ' h) '

41 4

i) ' 2 3 7 18 j) ' 1 5 3

y x y x x x

a xy y

x x a x

a x xy y

a x a x x

y yx x xw

y x x x y x x x

2.

2

3

2 2 2 2 sin

sin4 loga) b)

4 1

c) 2 cosec d) 2 cosx x x

n xedR dp

dx x dx x

dy dse e x x e

dx dx

3.

a) ' 1 b) ' 2 21.354

c) ' 1 27

f f

f



4.

3

2

33 2 22

22 2

2 2 2

3 4 2 4 2

a) 4cos .sin b) tan

c) ' 2cos 2 d) .cos sin

2sine) cos cos f)

1 cos

6 sin cos

g) 2 sec h) 2 tan sec

i) 24 tan 3 sec 3 j) ' cos 2 sin

dy dyx x x

dx dx

duf x x x x x

dx

dy xy x x

dx x

dyx x x

dx

dy dyx x x x

dx dx

dyx x x f x x x x x

dx

5.

2 2

2 2 2

a) 6 cos 4 b) 9 sin 2 3 cos 2 3

cos cosc) d)

2 2 sin

e) 6 cos .sin f) 2sin cos sin 2

dy dyx x x x

dx dx

dy x d x

dx dxx x

d dyx x x x x x

dx dx

You have reached the end of this learning unit, and so you should be able to

differentiate composite functions using the chain rule together with the rules for

combined expressions: constant times a function rule for sums and differences product rule quotient rule

We now move on to learning unit 7 on higher order derivatives.


DIFFERENTIATION Higher order

derivatives

CONTENTS PAGE

1. HIGHER ORDER DERIVATIVES ..................................................................... 71 2. RESPONSES TO ACTIVITY .............................................................................. 77

MODULE 6

LEARNING UNIT 7

OUTCOMES

At the end of this learning unit, you should be able to repeatedly differentiate a function to find higher order derivatives.

Module 6 Learning unit 7 DIFFERENTIATION: Higher order derivatives


1. HIGHER ORDER DERIVATIVES

The function dy

dx or 'f x is again a function of x on its own, which in turn may be

differentiated.

3 2For example if 5 , then 15dy

y x xdx

and 215x is a function that can be

differentiated with respect to x.

3 2Thus if 5 , then 15 30 .d dy d

y x x xdx dx dx

The expression [30x] is called the second differential coefficient (second derivative)

of the original function, and is denoted by the symbol2

2which means .

d y d dy

dx dxdx

2

2

2

2

Thus measures the rate at which is changing with respect to ,

just as measures the rate at which is changing with respect to .

Consequently can also be differentiated with re

d y dyx

dxdxdy

y xdx

d y

dx

3

3

spect to ,

and the result is the third differential coefficient (third derivative) of with respect to .

It is represented by the symbol .

x

y x

d y

dx

Thus in the above example:

3

2

2

2

3

3

4

4

Function: 5

First derivative: 15

Second derivative: 30

Third derivative: 30

Fourth derivative: 0

y x

dyx

dx

d yx

dx

d y

dx

d y

dx

As seen above, the repeated differential soon becomes 0 in some cases. In other cases it may continue indefinitely.



Example 1

Find the repeated differential coefficients of 3 2 1y x x x .

2

2

2

3

3

4

4

3 2 1

6 2

6

0, as well as all the higher differential coefficients

dyx x

dx

d yx

dx

d y

dx

d y

dx

Example 2

Find the repeated differential coefficients of 2

1 1y

x x .

1 2

2 3

23 4

2

34 5

3

1. 2.

2 6

6 24

y x x

dyx x

dx

d yx x

dx

d yx x

dx

Obviously this process will never end.

Notation for higher order derivatives

The following notations can be used: Function y f x y y x

First derivative dy

dx 'f x 'y D y ' x

Second derivative 2

2

d y

dx ''f x ''y D2 y '' x

Third derivative 3

3

d y

dx '''f x '''y D3 y ''' x

Fourth derivative 4

4

d y

dx v'f x 'vy D4 y 'v x



The successive differential coefficients are read as follows:

2

2

3

3

- by

- two by two

- three by three, etc.

dydy dx

dx

d yd y dx

dx

d yd y dx

dx

Note: There is a big difference between 22

2and .

d y dy

dxdx

There is also a big difference between 33

3 and

d y dy

dxdx

.

The next example will illustrate the difference:

Example 3

If 21y x x , determine 33

3 and

d y dy

dxdx

.

Solution

2

2

2

3

3

33

33

3

1

1 2

2

0

and 1 2

Clearly

y x x

dyx

dx

d y

dx

d y

dx

dyx

dx

d y dy

dxdx

Example 4

2

2Determine ''' if , simplify and give your answer with positive indices.

1f x f x

x

Solution You can use the quotient rule or rewrite f(x) to use the product rule.

122

22 1

1f x x

x



22

2222

3 22 2

3 22 2 2

2

3 22 2

2 2

32

2

32

2

32

' 2 1 1 2

44 1

1

'' 4 2 1 2 1 4 Using the product rule

16 1 4 1

16 4

1 1

16 4 1

1

12 4

1

4 3 1

1

f x x x

xx x

x

f x x x x x

x x x

x

x x

x x

x

x

x

x

x

2

32

3 22 2 2

232

3 22 2 2

52

22 2 2

62

2 2

42

3 3

42

4 3 1

1

1 4 6 4 3 1 3 1 2''' Using the quotient rule

1

1 24 24 3 1 1

1

1 1 24 24 3 1

1

1 24 24 3 1

1

24 24 72 24

1

x

x

x x x x xf x

x

x x x x x

x

x x x x x

x

x x x x

x

x x x x

x



3

42

2

42

48 48

1

48 1

1

x x

x

x x

x

Example 5

If 21 2 3u x x , determine an expression for 22

22

2 2d u d du

udx dxdx

.

Solution Find the values of the derivatives first, then substitute into the given expression and simplify.

2

2 2

2

2

2 6

2 6 4 24 36

6

dux

dx

dux x x

dx

d u

dx

22 2 2

2

2 2 3

2 3

1 2 3 2 1 2 3 2 6

4 12 1 2 3

4 8 12 12 24 36

4 20 36 36

d du x x x x x

dx dx

x x x

x x x x x

x x x

Now:

222

2

2 3 2

2 3 2

2 3

2 2

6 2 4 20 36 36 4 24 36 2

6 8 40 72 72 4 24 36 2

4 16 36 72

d u d duu

dx dxdx

x x x x x

x x x x x

x x x



ACTIVITY 1 1. Determine 'y in a) to d):

1

2

2

2

2

1a)

9

b)1

2c)

2 3d)

1

yxx

yx

xy

x

x xy

x

2

22. Determine if:

a)

b)2 1

d y

dxa bx

ya bx

xy

x

4 3 2

3. Determine ''' if:

a) 3 8 12 5

1b)

4

c)

f x

f x x x x

f xx

f x a bx

2

2

26 2

2

4. Find the values of ' and '' for the given values of :

a) ;

b) 9; 4

1 15. If 7 show that 6

2 10

y y x

ay ax x a

ax

y x x x

ds d ss t t t s

dt dt

6. 2

2If 3sin(2 3) show that 4 0

d yy x y

dx

7. If sin 2cos show that ''' '' ' 0y x x y y y y Remember to check the response on page 77.




Activity 1

12

52

32

3

2 2

2 2

2 2

2 3

2

2

1. a) '9

2b) '

2 2 1

2c) '

21 4

d) '1 2 3

42. a)

2b)

2 1

xy

x

xy

x

yx x

xy

x x x

d y ab

dx a bx

xd y

dx x

52

4

3

3. a) ''' 24 3 2

6b) '''

4

3c) '''

8

f x x

f xx

bf x

a bx

' 0

4. a) 1''

2

41'

5b) 4236

''125

yx a

ya

yx

y

5. 6

5

24

7

42

210

s t

dst

dt

d st

dt

Now left-hand side 2

22

1 1

2 10

ds d st t

dt dt



5 2 4

6 6

6

6

1 142 210

2 10

21 21

42

6 7

6

Right-hand side

t t t t

t t

t

t

s

6. 3sin 2 3y x

2

2

2

2

6cos 2 3

12sin 2 3

Left-hand side = 4

12sin 2 3 4 3sin 2 3

0

Right-hand side

dyx

dx

d yx

dx

d yy

dx

x x

7. sin 2cos

' cos 2sin

'' sin 2cos

''' cos 2sin

' '' ''' 0

y x x

y x x

y x x

y x x

y y y y

Learning unit 7 on higher order derivatives is now complete, so you should be able to

repeatedly differentiate a function to find higher order derivatives. The next learning unit 8 focuses on the applications of differentiation.


DIFFERENTIATION Applications I

CONTENTS PAGE

1. L’HOSPITAL’S RULE ........................................................................................ 80 2. THE GRADIENT OF A CURVE ......................................................................... 81 2.1 The equation of a tangent to a curve ..................................................................... 81 2.2 The equation of a normal to a curve ..................................................................... 83 3. RATE OF CHANGE ............................................................................................ 84 3.1 Velocity ................................................................................................................. 84 3.2 Acceleration .......................................................................................................... 84 4. RESPONSES TO ACTIVITIES ........................................................................... 86 4.1 Activity 1 .............................................................................................................. 86 4.2 Activity 2 .............................................................................................................. 87 4.3 Activity 3 .............................................................................................................. 87

MODULE 6

LEARNING UNIT 8

OUTCOMES

At the end of this learning unit, you should be able to

determine limits of the form 00 with ' 'Hospital s rule

determine the gradient to a curve find the equations of the tangent and the normal to a curve

Module 6 Learning unit 8 DIFFERENTIATION: Applications I


1. L’HOSPITAL’S RULE

Sometimes when we calculate the limit of a quotient we obtain the form 00 . In cases

where it is difficult to simplify the fraction we can use L’Hospital’s rule. L’Hospital’s rule states:

'0If lim is of the form then lim lim

0 '

'if lim is meaningful.

'

x a x a x a

x a

f x f x f x

g x g x g x

f x

g x

In some cases L’Hospital’s rule must be applied more than once until the denominator is not 0.

Example 1

0

0

tan 2Evaluate lim

tan 2tan 2 0

It is clear that the lim yields the indeterminate form .tan 2 0

x

x

x x

x xx x

x x

2

20 0

Therefore we apply L'Hospital's rule:

tan 2 1 2sec 2lim lim

tan 2 1 2sec 21 2

1 23

x x

x x x

x x x

Example 2 2 2

20 0

1 2Evaluate lim lim

2

t t t t

t t

e e e e

tt

00

2

0

which is still in the form apply rule again

4lim

23

2

t t

t

e e




a) 2

1

1lim

1x

x

x

b) 4 2

2

7 12lim

2x

x x

x

c) 2

20

sinlim

1 cos


2. THE GRADIENT OF A CURVE

From the definition of the derivative we know that the gradient (slope) of a curve

y f x at any point ,P a b is given by 'f a .

Example 3

Determine the slope of the curve 2y x at the point 3x .

3

Slope = 2 and 6.x

dy dyx

dx dx

Example 4

At which point on the curve 2

yx

is the gradient to the curve equal to –2?

The gradient = 22

22

dyx

dx x

Put the gradient = -2

2

2

2

22

2 2

1

1

x

x

x

x

2.1 The equation of a tangent to a curve

From the definition of the derivative we know that the gradient (slope) of a curve

y f x at any point ,P a b is given by 'f a and this is also equal to the gradient

of the tangent to the curve at point ,P a b .

The tangent to the curve is passing through ,P a b , thus

The equation of the tangent is: 'y b f a x a



Example 5

Find the equation of the tangent to the curve of y =3x2 + 4x + 5

at the point (3; 10). Solution Method 1

2

In this case 3 and 10.

Let 3 4 5

' 6 4

' 3 6 3 4

14

Equation of tangent is

10 14 3 Substitute values in 1

10 14 42

14 32

OR

14 32 0

a b

f x x x

f x x

f

y x

y x

y x

x y

Method 2 We can also use the gradient-intercept form of the straight line:

Gradient y-intercept

y mx c .

Let the equation of the tangent be y mx c . 2Now 3 4 5

6 4

y x x

dyx

dx

If 3, the gradient of the tangent: 6 3 4

14

x m

The tangent is passing through the point (3; 10), therefore these values must satisfy the equation.

10 14 3

10 42

32

Equation of tangent: 14 32

c

c

c

y x



2.2 The equation of a normal to a curve

The normal line to the curve y = f(x) at a point P (a, b) is defined as the line through P (a, b) perpendicular to the tangent at that point, in this case P (a, b).

Hence the gradient of the normal to y is equal to 1

'f a

because the conditions for two

lines to be perpendicular to each other is that the product of their gradients equals –1.

The equation of the normal at ,P a b is thus

Equation of the normal: 1

'y b x a

f a

Example 6

Find the equation of the normal to the curve of y =3x2 + 4x + 5

at the point (3; 10). Solution

2Let 3 4 5

' 6 4

' 3 6 3 4

14

Equation of normal is

110 3

143

1014 14

310

14 14143

14 14OR

14 143 0

f x x x

f x x

f

y x

xy

xy

x

x y

ACTIVITY 2 1. Determine the equation of (a) the tangent and (b) the normal to the

curve 2 2 25x y at the point (4 , 3) on the curve.

2. Find the equation of the tangent to the curve 2

2

1

xy

x

at the origin.




3. RATE OF CHANGE

We have said dy

dxis the instantaneous rate of change of y with respect to x for a

definite value of x.

For example, for the function 2y x we have 2dy

xdx

. When x = 4 then 8dy

dx .

The rate of change of y is therefore 8 units per unit change in x. The word instantaneous is often omitted, and it is clear that the rate of change of a function at a point x is equivalent to the gradient of the tangent to the function at the point x. When time is the independent variable, we get a very important application of the rate of change.

3.1 Velocity

Let a point P move along a straight line AB, and let s be the displacement measured from any point, say 0, on AB to P, and let t be the corresponding time taken. For every value of t, P will have a different position and therefore s will change. Therefore s is a function of t.

Thus s f t .

Now s

t

is the average velocity when P moves a distance s in the time t . In the

general case for any type of motion, the velocity at any instant at any point is defined as the limit of the average velocity when 0t .

Velocity ds

vdt

Thus velocity is the rate of change of displacement with respect to time.

3.2 Acceleration

2

2

Acceleration = Rate of change of velocity with respect to time

=dv

dtd ds

dt dt

d s

dt



Example 7

A body moves according to the equation 225s t t where s is the distance in metres and t the time in seconds from a reference point A. Find the velocity of the body at t = 3. Solution

225

Velocity 25 2

When 3, the velocity 25 2 3 19 m/s

s t t

dst

dtt

Example 8 The distance s metres moved by a body in t seconds is given by

3 22 13 24 10s t t t . Find a) the velocity when t = 4 seconds b) the value of t when the body comes to rest

c) the value of t when the acceleration is 10 m/s2 Solution

3 2

2

2

a) 2 13 24 10

Velocity 6 26 24

When 4, the velocity 6 4 26 4 24

96 104 24 m/s

16 m/s

s t t t

dst t

dt

t

2

2

43

13

b) When the body comes to rest 0

6 26 24 0

3 13 12 0

3 4 3 0

3

The values of when the body comes to rest are 3 seconds or 1 seconds.

ds

dt

t t

t t

t t

t or

t



2

2

2

2

2

c) Velocity 6 26 24

Acceleration 12 26

Acceleration is given as 10 m/s

12 26=10

12 =36

=3

The value of when the acceleration is 10 m/s is 3 seconds.

dst t

dt

d st

dt

t

t

t

t t

ACTIVITY 3 1. A missile is fired into the air. The height h in metres of the missile

after t seconds is given by 219.2 4.8h t t . Find a) the initial velocity of the missile (that is at t = 0) b) the height attained when its velocity is one-half of its initial velocity 2. A particle moving in a straight line is a distance s metres from a fixed

point after t seconds, where s = t3 4t2 + 5t. Find an expression for the velocity and the acceleration after t seconds. For what values of t is the particle stationary and what is the acceleration at these times?



4.1 Activity 1

a)

2

1 1

1 2lim lim

1 12 1

2

x x

x x

x

b)

4 2 3

2 2

3

7 12 4 14lim lim

2 1

4 2 14 2

32 28

4

x x

x x x x

x



c)

2

20 0

0

0

sin 2sin .coslim lim

2cos . sin1 cos

2sin .coslim

2cos .sinlim 1

1

4.2 Activity 2

1.

12

12

2 2

2 2

2

2

a) 25

25 25

125 2

2 254

Gradient of tangent at 4;3 :3

x y

y x x

dy x xx x

dx yxdy

dx

4Equation of tangent: 3 4

33 9 4 16

4 3 25 0

y x

y x

x y

43

1 3b) Gradient of normal

4

3Equation of normal: 3 4

44 12 3 12

3 4 0

y x

y x

x y

2. The equation of the tangent at the origin is 2y x

4.3 Activity 3

1. a) 19.2 m/s b) 14.4 m

2. 2

2

2

2 2

Velocity 3 8 5

Acceleration: 6 8

2Stationary after 1 and 1 seconds

3

Acceleration 2 m/s and 2 m/s

dsv t t

dt

d s dvt

dtdt



You have reached the end of this learning unit and you should be able to

determine limits of the form 00 with ' 'Hospital s rule

determine the gradient to a curve find the equations of the tangent and the normal to a curve

Learning unit 9 deals with the second part of applications, namely maxima and

minima.


DIFFERENTIATION Applications II: maxima and minima

CONTENTS PAGE

1. MAXIMA AND MINIMA ................................................................................... 89 1.1 Definitions............................................................................................................. 89 1.2 Maximum value .................................................................................................... 90 1.3 Minimum value ..................................................................................................... 90 2. DERIVED CURVES ............................................................................................ 91 3. PRACTICAL APPLICATIONS ........................................................................... 99 4. RESPONSES TO ACTIVITIES ......................................................................... 103 4.1 Activity 1 ............................................................................................................ 103 4.2 Activity 2 ............................................................................................................ 107

MODULE 6

LEARNING UNIT 9

OUTCOMES

At the end of this learning unit, you should be able to determine the maximum and minimum values of a function use the maximum and minimum values to sketch a graph

Module 6 Learning unit 9 DIFFERENTIATION: Applications II: maxima and minima


1. MAXIMA AND MINIMA

1.1 Definitions

Maximum A function reaches a maximum value when it stops increasing and starts decreasing. Minimum A function reaches a minimum value when it stops decreasing and starts increasing. The following sketch will make this clear. We use M to denote a maximum and m to denote a minimum as labelled in figure 1.

Figure 1 The sketch shows clearly that the slope of the tangent to the curve at the points M and m is 0. We know that the slope of the curve at any point on the curve is equal to the

derivative at that point. This means that 0dy

dx at point M and m.

If we solve the equation 0dy

dx , we can find the value of x where y is a maximum or

minimum. To distinguish between a maximum and minimum, the change of the curve in the vicinity of these points must be studied. We also refer to these values as extreme values, turning points or stationary points. Note that a maximum is not necessarily the highest value or a minimum the lowest value. These values are therefore sometimes called local maxima or minima.



1.2 Maximum value

Figure 2

It is clear in figure 2 that to the left of M, that is at A, the slope is positive, while to the right of M it is negative. The straight line underneath represents the approximate

shape of the slope curve near a value of x that makes 0dy

dx .

Now the slope of this new curve at this point is clearly negative, that is d dy

dx dx

is

negative when y reaches a maximum value. 2

2Thus 0 when is a maximum.

d yy

dx

1.3 Minimum value

In figure 3 it is clear that to the left of m, at C, the slope is negative, and to the right of m at D, it is positive. The straight line above the curve represents the approximate

shape of the curve of the slope near a value of x that makes 0dy

dx .

Now the slope of this new curve at the point is clearly positive, that is d dy

dx dx

is

positive when a minimum is reached. 2

2Thus 0 when is a minimum.

d yy

dx



Figure 3 The following procedure is used to find maximum or minimum values:

Put 0dy

dx to find the possible values of x.

Find 2

2

d y

dx. Substitute the values of x into

2

2

d y

dx.

If 2

20

d y

dx the function reaches a maximum at that point.

If 2

2 0 d y

dx the function reaches a minimum at that point.

2. DERIVED CURVES

Consider the graph of y f x in figure 4.

Figure 4 At point A the function reaches a maximum.



At point B the function reaches a minimum. At point C the function “appears” to bend, reach a half minimum/maximum. At this point the curve levels off and increases again immediately. This is an example of an inflection point. Derived curves are the graphs of derivatives. The curves are illustrated in figure 5.

Figure 5 The turning points of f (x) are at A, B and C.

The graph of the first derivative 'y f x shows clearly that 0dy

dx at the turning

points.



The graph of the second derivative ''y f x shows clearly that when

2

2

2

2

2

2

is a maximum, '' 0;

is a minimum, '' 0;

has an inflection point, '' 0.

d yy f x f x

dx

d yy f x f x

dx

d yy f x f x

dx

This information allows the value of the turning points to be determined by

differentiating the function and solving the equation 0dy

dx for x.

By substituting this value in f x , the corresponding y value is obtained.

The nature of the turning point (maximum, minimum or inflection point) can then be

determined by considering2

2

d y

dx.

We will use this information to sketch curves.

Example 1 Find the coordinates of the maximum and/or minimum turning points of the curve

defined by 3 26 9 2y x x x and sketch the curve. Solution

3 2

2

2

2

(1) Turning points occur at 0

6 9 2

3 12 9

For critical values 0

3 12 9 0

4 3 0

1 3 0

1 and 3

dy

dx

y x x x

dyx x

dxdy

dx

x x

x x

x x

x x

We need to find the nature of these points.

(2) To determine the nature of these points substitute 1 and 3x x into 2

2

d y

dx.



22

23 12 9 Thus 6 12

dy d yx x x

dx dx .

2

2

2

2

6 12

At 1: 6 1 12

6 0

negative

d yx

dx

d yx

dx

Thus a maximum value at x = 1. To find the maximum value substitute x = 1 in the original equation.

3 2max 1 6 1 9 1 2

1 6 9 2

2

y

Thus the coordinates of the maximum turning point are (1;2).

2

2At 3 : 6 3 12

6

positive

d yx

dx

Thus a minimum value at x = 3. To find the minimum value substitute x = 3 in the original equation.

3 2min 3 6 3 9 3 2

27 54 27 1

2

y

Thus the coordinates of the minimum turning point are (3; -2).

We are ready to sketch the curve:

Figure 6



Example 2

Find the turning points of the graph of 2

23

3 1xf x

x

and establish which is the

minimum and maximum. Sketch the graph. Solution

2

23

223

223

2 2

223

2

223

3 1

6 3 1 1'

6 4 3 1

3 4 1

xf x

x

x x xf x

x

x x x

x

x x

x

2

223

2

For critical values put ' 0

3 4 10

23 4 1 0 Provided that 0

3

3 1 1 0

11 and

3

f x

x x

x

x x x

x x

x x

The nature of the turning points:

2

223

2 22 23 3

2223

3 4 1'

6 4 3 4 1 2 1''

x xf x

x

x x x x xf x

x

213

413

min 13

2 0At 1: '' 1 0 positive

Minimum value at 1

3 1 2 31

1 1

6

Minimum turning point 1; 6

x f

x

y f



21 1 4 13 3 3 3

213

29

19

3 11 9

max 13

2313

2 1 21At : ''

3

0 2 9

9 1

2

0 negative

1Maximum value at

311

3

2 3

3 1

2

1Maximum turning point ; 2

3

x f x

x

y f

Note: 23 is undefined when .f x x Points like these need further investigation as

shown in figure 7.

Figure 7



Example 3 Find the coordinates of the maximum and/or minimum turning points of the curve

. xy x e . Solution

. 1 1

1

x

x x

x

y xe

dyx e e

dx

e x


1 0

0; 1

x

x

dy

dx

e x

e x

Nature of the point

2

21 1 1

1 1

2

x x

x

x

d ye x e

dx

e x

e x

21

2

1min

If 1: 1 2

11

0 positive

Minimum value when 1

1

1

0,368

Coordinates of minimum turning point 1; 0,368

d yx e

dx

e

x

y e

e

Note: The expression xe is always positive for any value of x. Thus 0.xe

Also, it is not necessary to find the actual value of 2

2

d y

dx.



Example 4

Find the maximum and/or minimum values of 2 1

xy

x

and sketch the graph.

Solution

2

2

22

2

22

1

1 1 2

1

1

1

xy

x

x x xdy

dx x

x

x

Turning points

2

22

2

2

10

1

1 0

1

1

x

x

x

x

x

The point to note is that the denominator 22 1x can be ignored in solving 0dy

dx as

it may never equal 0.

There is no value we can give to except 1 or 1 that will make 0.dy

x xdx

Nature of the turning points

22 2 22

2 42

22

2 4

22

2 4

1 2 1 2 1 2

1

2 2 0 negative valueIf 1, negative value

positive value2

a maximum.

2 2 0 positive valueIf 1, positive value

positive value2

a minimum.

x x x x xd y

dx x

d yx

dx

y

d yx

dx

y

1 1

Maximum value of 1 1 2

1 1Minimum value of

1 1 2

y

y

Sketch:



Figure 8

ACTIVITY 1 1. Find the maximum and/or minimum turning points of the curve

25 1y x x .

2. Determine the coordinates of the maximum and/or minimum points of

the curve x xy e e . 3. Find the coordinates of the maximum and minimum points of the curve

defined by 212 10 .y n x x x 4. Find sin cos for 0 .y x x x

5. Sketch the curve of 1 2 3y x x x

6. Sketch 3 22 11 12f x x x x .

7. Sketch the curve 3 3 14y x x . Remember to check the response on page 103.

3. PRACTICAL APPLICATIONS

Example 5 Find the dimensions of the largest open container that can be made from a sheet of metal 60 cm by 28 cm, by cutting equal squares from the corners and turning up the sides.

Let the side of a square = x cm. We want a relation between the volume (V) of the container and x.



2 3

2

Height of container

Width of container 28 2

Length of container 60 2

Volume of container 28 2 60 2

1 680 176 4

1 680 352 12

x cm

x cm

x cm

V x x x

x x x

dVx x

dx

2

2

2

2


12 352 1 680 0

3 88 420 0

3 70 6 0

706 and

3

352 24

dV

dx

x x

x x

x x

x

d Vx

dx

2

2If 6 : 352 24 6

352 144

208

= Negative

Maximum when 6

d Vx

dx

x

Dimensions of largest open container:

Height = 6

Width = 28 2 6 16

Length = 60 2 6 48

cm

cm

cm

Example 6

2

120The cost of manufacture , per km of an electric cable is given by 600 ,

where is its cross-section in cm . Find:

Rc c xx

x

a) the cross-section for which the cost is least

b) the least cost per km Solution



2

120600 ............................. 1

120600

c xx

dc

dx x

2

2

2


120600 0

600 120

120

600

0.2

0.447

dc

dx

x

x

x

x

The negative root has no meaning in this connection, and is disregarded. 2

2 3

240d c

dx x

2

2 3

240When 0.447;

0.447

0 positive

d cx

dx

Minimum when x = 0.447

a) Cost is a minimum for a cross-section of 0.447 cm2

b) Substituting for x = 0.447 in (1), we get the minimum cost.

120Thus R 600 0.447

0.447R 536.66

c

Example 7 An open tank is to be made of sheet iron; it must have a square base and sides perpendicular to the base. Its capacity is to be 8 m3. Find the side of the square base and the depth, so that the least amount of sheet iron may be used. Solution Let the length of the square base = x cm and the height of the tank = h cm. We need a relation between x and the surface area (least amount of sheet iron). Surface area: A = Area of base + Area of sides

2 4 ................................... 1A x xh

[Note: If the tank is closed it will be 2 area of base + area of sides ]

We must now find a relation between x and h because in (1) we have three variables.

The capacity is given as 8 m3.



2

2

22

2

8

8substitute in 1

4 8.

132

Now we have two variables only

x h

hx

xA x

x

xx

2

2

3

3

3

322


322 0

2 32

16

16

2.52

dAx

dx xdA

dx

xx

x

x

x

Nature of values:

2

2 3

2

2 3

2

642

64If 2.52 : 2

2.52

0 positive

Minimum when 2,52

8

2.52

1.26

d A

dx x

d Ax

dx

x

h

To use the least amount of sheet iron the side of the square base must be 2.52 m and the height of the tank 1.26 m.



ACTIVITY 2 1. A rectangular field is to be enclosed by a fence and divided into smaller plots by a fence parallel to one of the sides. Find the dimensions of the largest such field which can be enclosed by 1 200 m of fencing. 2. The distance s (in metres), travelled by a body projected vertically upward in time t seconds, is given by the formula s = 120t 16t2 . Find the greatest height which the body will reach and the time taken. Remember to check the response on page 107.


4.1 Activity 1

1. Max 5;0 ; Min 1; 32

2. Min 0;2

3. Max 2; 7.68 ; Min 3; 7.82

The next answers are given in more detail. 4. sin cosy x x

cos sindy

x xdx

Turning points: cos sin 0

sin cos

sin1

cos

that is tan 1

4

x x

x x

x

x

x

x

Nature of points

2

2

2

2

Now: sin cos

1 1If then negative

4 2 2 is a maximum

d yx x

dx

d yx

dx

y



Maximum value of sin cos4 4

2

2

2

y

5. Remove brackets to find

3 24 6y x x x Turning points

23 8 1dy

x xdx

2

Let 0 for extreme values

3 8 1 0

8 64 12

68 7, 21

62.54 or 0.13

dy

dx

x x

x


2

2

2

2

3 2

6 8

If 2.54 : then 15.24 8

7.24

0

Graph has a minimum at 2.54

Value of 2.54 4 2.54 2.54 6

16.39 25.81 2.54 6

0.88

d yx

dx

d yx

dx

x

y

2

2

3 2

If 0.13: then 6 0.13 8

0.78 8

7.22

0

Graph has a minimum at 0.13

Value of 0.13 4 0.13 0.13 6

0.0022 0.0676 6.13

6.06

d yx

dx

x

y



Therefore A is the point (0.13; 6.06) and B is the point (2.54; 0.88) in figure 9. Intercepts: y-intercept: y = 6 x-intercepts: We can find these easily as the function is given in factor

from 1 2 3y x x x

Thus x = , x = 2 or x = 3. If the function is not given in this form, we may omit finding the exact

values of the x-intercepts.

Figure 9: 3 24 6y x x x

f) 3 22 11 12f x x x x

2' 3 4 11

'' 6 4

f x x x

f x x

Turning points

2

Solve ' 0

3 4 11 0

4 16 132

64 12.17

616.17 8.17

or 6 6

2.70 or 1.36

f x

x x

x




3 2

For 2.70 '' 6 2.70 4 0

2.70 is a point of maximum value.

Value of 2.70 2 2.70 11 2.70 12

19.68 14.58 29.70 12

12.6

x f x

x

y

3 2

For 1.36 '' 6 1.36 4 0

1.36 is a point of minimum value.

Value of 1.36 2 1.36 11 1.36 12

2.52 3.70 14.96 12

20.74

x f x

x

y

y-intercept: y = 12 Sketch:

Figure 10

g) 3 3 14y x x y-intercept = 14



Turning points

2

2

3 3

has no real roots since 1 has no real solution.

dyx

dxdy

xdx

Therefore there are no turning points. Points of inflection

2

26

0 is a point of inflection.

d yx

dxx

Large values of x:

If x then y

If x then y Sketch:

Figure 11

4.2 Activity 2

1. Let the width be x metres.

1 200 3

Length 2

3600

2

x

x



2

3Area of field: 600

2

3600

2

600 3

A x x

xx

dAx

dx

2

2


600 3 0

3 600

200

3

0 negative

Maximum when 200

dA

dxx

x

x

d A

dx

x

3 200

Length 6002 1

300

Dimensions of largest field

Length = 300 m

Width = 200 m

2. 2120 16s t t

Maximum height will occur where 0ds

dt .

2

2

0

120 32 0

32 120

3,75 seconds

120 16

120 3,75 16 3,75

225 metres

ds

dtt

t

t

s t t

Check with second derivative test that it is a maximum:

2

232 0 Maximum

d s

dt



You have now reached the end of this learning unit and you should be able to determine the maximum and minimum values of a function use the maximum and minimum values to sketch a graph

Next is the post-test on differentiation. Answer the questions first and then check your answers.


POST-TEST: DIFFERENTIATION

1. Given that 2( ) 4 3 2f x x x , find the values of a) (2) (1)f f b) (3 )f a c) (3 ) (3)f a f

d) (3 ) (3)f a f

a

2. Given 2 2( ) 5 and ( )

2

xg x x x f x

x

, determine

a) 0g

b) 0f

c) if 2 0a g a f

d) if 0y y f y

3. Calculate the following limits:

9

2

2

24

4 4

2 2

2

40

a) lim 3

3 7b) lim

1 4

2 8c) lim

5 4

d) lim

5 3e) lim

7 5

y

t

x

x a

x

y

t t

t t

x x

x x

x a

x a

x x

x x

2

3

8 2 3f) lim

2 3 13 5

g) lim2 3

x

x

x x

x xx

x

4. Find the following one-sided limits:

2 2

1 1lim and lim

2 2x xx x

5. Determine the derivative of 3 23 2 1f x x x x from first principles.

6. Find dy

dx from first principles if

2

2y

x .

7. 2 2Find if 10 .da

a b x bx xdb

Module 6 DIFFERENTIATION: Post-test


8. Determine the gradient of the curve x = 2y2 at the point where x = 8.

9. Determine dy

dx, simplify and write the answers with positive indices and in

surd (root) form.

324

3 3

2

a) 1

b) 1

c) 1

y x

y x

y x x

10. 3

4

22

6If 6 determine .

dyy t

dtt

11. At what point (coordinates) on the curve of 2

3 13

y x is the gradient of

the curve equal to 1?

12. 2 3

Determine the gradient of the tangent to the curve at the point 5

on the curve where 4. Give your answer in simplified surd form.

tf t

t

t

13. Differentiate with respect to x, and simplify your answers to a single term,

with positive indices and in surd form where possible:

2

2

2

a) 1 2 2 b)

1 1c) d)

1 4

1e)

1

qpx x x y x ax b

x xy y

x x

xr

x

14. Determine the gradient of the tangent to the curve 23 1

+5

xf x

x

at the

point 1

1;3 2

. Give your answer in simplified surd form.

15. Find the x value(s) of the point on the curve 2

4

xy

x

where the gradient of

the curve equals 0. (Hint: 4 0x ).

16. 32If 1 , determine ''' . Simplify your answer to a single term,

and in surd form.

f x x f x

Module 6 DIFFERENTIATION: Post-test solutions


17. Differentiate and simplify the answer to a single fraction if

3

42n

2x x

y ex

18. Differentiate with respect to the independent variable and simplify:

2

3 2

a) 1 .sin 3

b) sin 2

y x x

t t

c) sin . tan sec 1

1d) tan

1

y x x x

x

x

19. Determine the following:

0

20

3 2

21

2 2a) lim

sinb) lim

2 4 3c) lim

4 5 1

h

x

x

h

h

x x

x

x x x

x x

20. The displacement s metres of a particle from a fixed point at a time t seconds

is given by 1010 5t

s te

. Find expressions for the a) velocity at time t b) acceleration at time t c) value of t at which the velocity is 0

21. A tank of water is filled in such a way that at the end of t minutes there are 3

2

3

tt litres of water in the tank. The person filling the tank is instructed to

turn off the water when the water is entering the tank at a rate of 15 litres per minute. When should he turn off the water?

22. Find the coordinates of the maximum and/or minimum turning point(s) of the

curve defined by 4 28 16y x x . 23. Determine the coordinates of the maximum and/or minimum turning point(s)

of 1f x x

x and sketch the curve.

24. If 2 3 xf x x e determine the minimum value correct to two decimal

places.



25. Find the maximum value of sin 2y x x by using differential calculus for 0 x .

26. A rectangular box with a square base and open top is to be made. Find the

volume of the largest box that can be made from 1 200 cm2 of material. 27. The displacement x metres of an oscillating mechanism after a time t seconds

is given by 2,4sin 5 0,1x t metres.

Find the a) maximum displacement and the time at which it occurs b) displacement and velocity after 0,2 seconds

28. The motion of a particle performing damped vibrations is given by

sin 2ty e t , where y is its displacement from the mean position at time t. Determine the maximum value of y correct to three significant figures.

29. Find the maximum value of the function 21 2x x .

30. The efficiency of a screw is given by 24 3

3 4

t tE

t

where t is the angle of

pitch of the screw. Show that maximum efficiency is 25%.

31. Determine the maximum and minimum values of the function 3

3 93

xx and

give a rough sketch of the curve 3

3 93

xy x .

32. If 3 33 6x xy e e show that 2

29

d yy

dx .

33. Sketch the following curves:

3 2

2

a) 2 3 12

b) 1 2 3

y x x x

y x x x



DIFFERENTIATION: POST-TEST SOLUTIONS

1. 2( ) 4 3 2f x x x

a) 2(2) 4 2 3(2) 2

16 6 2

12

f

2(1) 4(1) 3(1) 2

4 3 2

3

12(2) (1)

3

4

f

f f

b) 2

2

2

2

(3 ) 4(3 ) 3(3 ) 2

4(9 6 ) 9 3 2

36 24 4 9 3 2

4 21 29

f a a a

a a a

a a a

a a

c) 2

2

(3) 4(3) 3(3) 2

4(9) 9 2

29

(3 ) (3) 4 21

f

f a f a a

d) 2(3 ) (3) 4 21

(4 21)

4 21

f a f a a

a a

a a

a

a

2. 2 2( ) 5 ( )

2

xg x x x f x

x

2a) (0) (0) 5(0) 0

2 0 2b) (0) 1

0 2 2

g

f



2

2 04

2

2

2

2

c) ( ) (2) 0

2 2( 5 ) 0

2 2

( 5 ) 0

5 0

5 0

0 or 5

d) 0

20

2

2 2 0

2 2 0

3 2 0

3 (3) 4(1)( 2)

2

0,56 3,56

g a f

a a

a a

a a

a a

a a

y f y

yy

y

y y y

y y y

y y

y

or

3.

9a) lim 3

12

2 3

yy

2

3 7b) lim

1 4

2 3 2 7

2 1 2 4

5 9

1 6

45

6

t

t t

t t



2

24

4

2 8c) lim

5 44 2

lim4 1

4 2

4 16

32

x

x

x x

x xx x

x x

4 4

2 2

2 2

2

d) lim

lim1

2

x a

x a

x a

x a

x a

a

2

40

5 3 3e) lim

57 5x

x x

x x

2 3

2 3

2

3

8 32

3 1

5

3

8 2 3f) lim

2 3 1

lim2

0

3 3g) lim

2 2

x

x x x

xx x

x

xx

x x

x x

2

2

14. lim

21

lim2

x

x

x

x

5. We must evaluate

0limh

f x h f x

h

.

3 2

3 2

3 2 2 3 2 2

2 2 3 2

3 2 1

Step 1: 3 2 1

3 3 3 6 3 2 2 1

Step 2: 3 3 6 3 2

f x x x x

f x h x h x h x h

x x h xh h x xh h x h

f x h f x x h xh h xh h h



2

2

2

0 0

2

2

3 3 6 3 2Step 3:

3 3 6 3 2

Step 4: lim lim 3 3 6 3 2

3 6 2

Thus ' 3 6 2

h h

h x xh x hf x h f x

h h

x xh x h

f x h f xx xh x h

h

x x

f x x x

6. We must evaluate

0limh

f x h f x

h

.

22

2

2

2 3 4 2

2 3 2 4

3 4 2

2 Put 2

Step 1: 2

Use the binomial theorem on

2 32 2 .....

1.2

2 4 6 .....

Step 2: 4 6 .....terms containing higher orders

y f x xx

f x h x h

x h

f x h x x h x h

x x h h x

f x h f x x h x h

3

3

0 0

3

3

of

Step 3: 4 6 terms containing higher orders of

Step 4: lim lim 4 6 terms containing higher orders of

4

(all terms containing become 0)

Thus 4

h h

h

f x h f xx xh h

h

f x h f xx xh h

h

x

h

dyx

dx

7. 22da

bx xdb

8.

1

2222

xx y y

Note that

1=

dxdy

dy

dx.



12

4

1

4

1 41

24

2

2

4

dxy

dy

dy

dx y

x

x

x

8

8

Gradient at 8 is

2

4 8

2

4.2. 2

1

8

x

x

dyx

dx

dy

dx

9. 3

432 24a) 1 1y x x

14

14

2

2

4 2

31 2

4

61

4

6

4. 1

dyx x

dx

xx

x

x

13

23

23

3 3

3

3 2

2 3

2

233

b) 1

1

11 3

3

1

1

y x

x

dyx x

dx

x x

x

x



12

12

2

2

2

2

c) 1

1

11 2 1

2

2 1

2. 1

y x x

x x

dyx x x

dx

x

x x

10. 3 3

4 42 2 2 26 6 6 6y t t t t

14

13

14

13

14

2 2 3

2 2

22

3

242

36 6 . 12 12

4

9

6 6

9

16

19

16

dyt t t t

dt

t t

t t

t t

tt

tt

tt

11. 1

22 23 1 3 1

3 3y x x

12

12

12

13 1 .3

3

13 1

3 1

1

3 1

dyx

dx

xx

x

1Given 1 thus 1

3 1

3 1 1

3 1 1

2

3

dy

dx x

x

x

x

Substitute the value of x into the original equation to find the value of y.



2 23 1

3 3

22 1

3

2

3

y

12. Gradient = 23

81 13.

13.

122 2

22

2

2 2

2

2

2

1a) 2 2 1 2 2 2 2

2

12 2

2 2

2 2 2 1

2 2

2 4 3

2 2

dx x x x x x

dx

xx x

x x

x x x x

x x

x x

x x

11

1 1

b)q qp p

q p p

dypx ax b x q ax b a

dx

ax b px ax b qax

2

2

1 1 1 1c)

1

2

1

x xdy

dx x

x

1 12 2

12

2 2 212

2

2 2

2 2

3

2 2

4 2 1 4 2d)

4

4 2 1

4 4

7

4 4

x x x x xdy

dx x

x x x x

x x

x x

x x



1 12 21 1

2 21 1 1 1

1e)

1 11 11 12 2

2 1 2 11

1 11 11 1 12 2

1 2 1 2 1

1 11 11 2 21 12 1

1 1112 1

21 1

11

21 1 1

11 1

x x x x

x

dr

dx

x xx x

x

x xx x x

x xx xx

x x

xx x

x x x

x x

14.

1 12 22 21

2

2

212

2 2

2 2

5 3 1 6 3 1'

5

5 6 3 1

5 3 1

15 1

5 3 1

x x x xf x

x

x x x

x x

x

x x

2 2

15 1 1Gradient

1 5 3 1 1

16

36 2

2 2

9

15. 2 12' 1 4 2 4

' 0

y x x x x

y



2 12

2

2 2

2

4 2 4 0

2 4 0

2 8 0

8 0

8 0

0 or 8

x x x x

x x x

x x x

x x

x x

x

16.

32

12

12

2

3' 1 2

2

3 1

f x x x

x x

1 12 2

1 12 2

12

12

1 12 2

32

32

32

2 2

2 2 2

2 2

2

2

2

2 2 212

2

2 2

2

3 3

2

3

2

3

2 2

1'' 3 1 3 1 2

2

3 1 3 1

3 1 3

1

6 3

1

1 12 6 3 1 2'''

1

1 12 6 3

1

12 12 6 3

1

9 6

1

9 6

1 1

f x x x x x

x x x

x x

x

x

x

x x x x xf x

x

x x x x

x

x x x x

x

x x

x

x x

x x

17. 32 2

4y x n e n x n x



2

2

2

3 1 1'

4 2 2

3 31

4 2 4 2

4 4 3 2 3 2

4 2 2

1

4

y n ex x

x x

x x x

x x

x

x

18.

12

1 12 2

12

12

1 12 2

12

2

2

2 2

2

2

2

2

2

2

a) 1 .sin 3

1 .sin 3 Product rule

11 .cos3 3 sin 3 1 2

2

3 1 .cos 3 sin 3

1 1

3 1 cos3 sin 3

1

3 1 cos3 sin 3

1

y x x

x x

dyx x x x x

dx

x x x x

x

x x x x

x

x x x x

x

3 2

32

3 12 2 2

22 2

22 2

2 2 2

b) sin 2

sin 2 Function of a function rule

3 sin 2 . sin 2 . 2

3 sin 2 .cos 2 . 2 2

3 2 sin 2 .cos 2 1

6 1 sin 2 .cos 2

t t

t t

d d dt t t t t t

dt dt dt

t t t t t

t t t t t

t t t t t



2c) cos tan sin sec sec tan

1 1sin tan tan

cos cos

sin

dyx x x x x x

dx

x x xx x

x

22

22

1 1 11d) sec .

1 1

2 1sec

11

x xd x

dx x x

x

xx

19. 0

2 2a) lim

h

h

h

12

12

12

0

12

0Substitution gives , use 'Hospital's rule:

0

2 0lim

1

(2)

1

2 2

h

h

20

sinb) lim

x

x x

x

0

0

0Substition gives use 'Hospital's rule:

0

1 cos= lim

2

0Substition gives use 'Hospital's rule again:

0

0 sin= lim

2

0

2

0

x

x

x

x

x

3 2

21

2 4 3c) lim

4 5 1x

x x x

x x



2

1

0Substition gives use 'Hospital's rule:

0

3 4 4= lim

8 5

3 4 4

8 5

3

3

1

x

x x

x

10

10 10

10

10 10

10 10

10

2

2

20. a) 5 Use product rule

15 5

10

15 1

10

1 1b) 5 . 1 5

10 10 10

1 15 5

10 100 10

1 15

10 100 10

t

t t

t

t t

t t

t

ds dte

dt dt

e t e

e t

d s te e

dt

te e

te

101

5100 5

1c) 1 0

10

10

10

10

t te

t

t

t

32

2

2

2

21.3

2

2 15

2 15 0

5 3 0

5

tV t

dVt t

dt

t t

t t

t t

t



4 2

3

3

2

22. 8 16

' 4 16

4 16 0

4 4 0

0; 2; 2

y x x

y x x

x x

x x

x x x

To find maximum and minimum do second derivative test.

2'' 12 16

0 : '' 16 0 maximum

2 : '' 8 0 minimum

2 : '' 4 0 minimum

y x

x y

x y

x y

2

2

2

2

2

3

123.

1' 1

11 0

10

1 0

1

2

2''

1: '' 0 min

1: '' 0 max

f x xx

f xx

x

x

x

x

x

y

f xx

x f x

x f x



2

2

2

24. 3

' 3 2

2 3

x

x x

x

f x x e

f x e x e x

e x x

2

1

' 0

3 1 0

3 of/or 1

'' 2 3 2 2

1: '' 0 min

Min 1 1 3 5, 44

x x

f x

x x

x x

f x e x x e x

x f x

f e

0,5235

25. sin 2

' 2cos 2 1

' 0

2cos 2 1 0

1cos 2

22 1.0472

0.5235

'' 4sin 2

'' 3.464

0 max

sin 2 0.5235 0.5235

0.023

x

y x x

y x

y

x

x

x

x

y x

y

y



26.

2 2

2

2

Volume = and Area = 4

1200 4

1200

4

x h x xh

x xh

xh

x

2 32 1200

3004 4

x xV x x

x

2

2

2

2

2

2

220

3

3300

4

For maximum put 0

3300

4

400

20

6Check with 2nd derivative test:

4

0 Thus a maximum value at 20

1200 400Volume of largest box = 400

80

4000 cm

x

dV x

dxdV

dx

x

x

x

d V x

dx

d Vx

dx



2

27 a) 2, 4 cos 5 0,1 5

12 cos 5 0,1

5 0,12

0,1

50.334

Maximum displacement 2.4sin 5 0.334 0.1

2.4

dxt

dtt

t

t

x

2

2

0.334

dCheck for max: 60sin 5 0.1 Remember the angle is in radians

d60 0 maximum

t

xt

dt

x

dt

b) 2.4 sin 1 0.1 12 cos 5 0.1

2.4 sin 0.9 12 cos 1 0.1

1.88 m = 7.45 m/s

dxx t

dt

0.554

28. ' sin 2 2 cos 2

sin 2 2cos 2 0

2cos 2 sin 2

tan 2 2

2 1,107

0.554

max: sin 2

sin 2 0.554

0,5139

t t

t

t

y e t e t

e t t

t t

t

t

t

y e t

e

Remember second derivative test.



2

2

2

29. 1 2

' 2 1 2 2

2 2 2 2

2 3 4

4' 0 2 and

3'' 3 4 2 3

6 10

'' 2 2 0 max

2 2 1 2 2

0

F x x x

F x x x x

x x x

x x

F x x x

F x x x

x

F

F

2

2

2

2 2

2

4 330.

3 4

3 4 4 6 4 3 4

3 4

12 2 24 16 12

3 4

t tE

t

t t t tdE

dt t

t t t t

t

2

2

2

2

12 18 12

3 4

6 2 3 2

3 4

t t

t

t t

t



2

2 2

4

2

3

3

21 12 2

12

54

' 0

2 3 2 0

2 1 2 0

1of/or 2

2

3 4 24 18 6 2 3 2 3 4 2 4''

3 4

3 4 24 18 48 2 3 2

3 4

150

3 4

1: '' 0 max

2

4 3

3 4

1

5 425%

E

t t

t t

t t

t t t t tE

t

t t t t

t

t

t E

E

3

2 2

31. 3 93

' 3 ' 0 3 3

'' 2

xf x x

f x x f x x x

f x x

3

3

3 3'' 0 min 3 3 9 9 2 3

3

3 3'' 0 ax 3 3 9 9 2 3

3

x

x

f x y

f x m y



3 3

3 3

3 3

32. ' 9 18

'' 27 54

9 3 6

9

x x

x x

x x

y e e

y e e

e e

y

3 2

2

2

2

3 2

2

33. a) 2 3 12

6 6 12

6 6 12 0

2 0

2 1 0

2 1

20 7

0 : 2 3 12 0

2 3 12 0

y x x x

dyx x

dx

x x

x x

x x

x x

y y

y x x

x x x

3 105

0 :4

1.81 3.31

x x

x x



2

2 2

2

b 1 2 3

Intercepts : axis: 6

axis : 1, 2, 3

Extreme values

' 2 1 2 3 1 3 1 2

1 2 2 3 1 3 1 2

1 4 5 11

Put ' 0

Turning points at 1, 1.15 or 2.40

y x x x

y y

x x

f x x x x x x x x

x x x x x x x

x x x

f x

x x x

When we take note of our mistakes we can avoid them in future. Before moving on take some time to write down mistakes you should avoid when you differentiate. We have concluded module 6 on differentiation and will now move to module 7, which deals with integration. We start with learning unit 1, the reverse of differentiation.

MAT1581 Mathematics I (Engineering) 135

M O D U L E 2

M O D U L E 7

INTEGRATION CONTENTS PAGE

LEARNING UNIT 1 REVERSE OF DIFFERENTIATION I 137

1. THE PROCESS OF INTEGRATION ................................................................. 138 2. THE GENERAL SOLUTION OF INTEGRALS OF THE FORM nax ............. 139 3. RESPONSES TO ACTIVITIES .......................................................................... 141 3.1 Activity 1 ............................................................................................................. 141 3.2 Activity 2 ............................................................................................................. 142 3.3 Activity 3 ............................................................................................................. 143

LEARNING UNIT 2 REVERSE OF DIFFERENTIATION II 144

1. SIMPLIFY THE INTEGRAND BEFORE INTEGRATION ............................. 145 2. DETERMINE THE VALUE OF THE CONSTANT OF INTEGRATION ....... 147 3. APPLICATION: DISPLACEMENT, VELOCITY, ACCELERATION ............ 148 4. RESPONSES TO ACTIVITIES .......................................................................... 149 4.1 Activity 1 ............................................................................................................. 149 4.2 Activity 2 ............................................................................................................. 150 4.3 Activity 3 ............................................................................................................. 150

LEARNING UNIT 3 METHOD OF SUBSTITUTION 152

1. METHOD OF SUBSTITUTION ........................................................................ 153 2. RESPONSE TO ACTIVITY ............................................................................... 156

LEARNING UNIT 4 STANDARD INTEGRALS 161

1. TABLE OF STANDARD INTEGRALS ............................................................ 162

2. INTEGRALS OF THE FORM

f (x)dx

f(x) .......................................................... 163

3. INTEGRALS OF THE FORM f xf ' x e dx .................................................. 165

4. INTEGRALS OF THE FORM f(x)f (x).a dx .................................................. 166

5. RESPONSES TO ACTIVITIES .......................................................................... 168 5.1 Activity 1 ............................................................................................................. 168 5.2 Activity 2 ............................................................................................................. 168 5.3 Activity 3 ............................................................................................................. 169 5.4 Activity 4 ............................................................................................................. 169

Module 7 INTEGRATION: Contents

MAT1581 Mathematics I (Engineering)

136

LEARNING UNIT 5 PARTIAL FRACTIONS 172

1. INTEGRATION USING PARTIAL FRACTIONS ............................................ 173 2. RESPONSE TO ACTIVITY ............................................................................... 175

LEARNING UNIT 6 TRIGONOMETRIC FUNCTIONS 178

1. INTEGRATION OF TRIGONOMETRIC FUNCTIONS .................................. 179 2. RESPONSES TO ACTIVITIES .......................................................................... 182 2.1 Activity 1 ............................................................................................................. 182 2.2 Activity 2 ............................................................................................................. 184

LEARNING UNIT 7 THE DEFINITE INTEGRAL 187

1. DEFINITION ...................................................................................................... 188 2. EVALUATION OF A DEFINITE INTEGRAL ................................................. 188 3. RESPONSE TO ACTIVITY ............................................................................... 190 4. EXERCISE 1 ....................................................................................................... 191

LEARNING UNIT 8 AREAS 193

1. INTRODUCTION ............................................................................................... 194 2. SUMMATION AND THE DEFINITE INTEGRAL .......................................... 195 3. THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE ................. 196 4. RESPONSES TO ACTIVITIES .......................................................................... 205 4.1 Activity 1 ............................................................................................................. 205 4.2 Activity 2 ............................................................................................................. 206 5. EXERCISE 1 ....................................................................................................... 208

POST-TEST 210

POST-TEST SOLUTIONS 212


INTEGRATION Reverse of

differentiation I

CONTENTS PAGE

1. THE PROCESS OF INTEGRATION ................................................................. 138 2. THE GENERAL SOLUTION OF INTEGRALS OF THE FORM nax ............. 139 3. RESPONSES TO ACTIVITIES .......................................................................... 141 3.1 Activity 1 ............................................................................................................. 141 3.2 Activity 2 ............................................................................................................. 142 3.3 Activity 3 ............................................................................................................. 143

MODULE 7

LEARNING UNIT 1

OUTCOMES

At the end of this learning unit, you should be able to explain what is meant by an indefinite integral explain what is meant by the term "integrand" explain what is meant by a constant of integration find solutions to integrals of the form nax dx

Module 7 Learning unit 1 INTEGRATION: Reverse of differentiation I


138

1. THE PROCESS OF INTEGRATION The process of integration reverses the process of differentiation. In differentiation we start with the function f(x) and then find the differential coefficient f'(x). For example:

3 2If , then ' 3 .f x x f x x Thus the integral of 3x2 is x3, and therefore integration

is the process of moving from f'(x) to f(x). The function being integrated is called the integrand. Thus 3x2 is the integrand in the example. Integration is also a process of summation or adding parts together and an elongated S, shown as , is used to replace the words "the integral of". The symbol, , is known as the integral sign. Thus from the example above 2 33 is x x .

In differentiation, the differential coefficient dy

dx indicates that a function of x is being

differentiated with respect to x, the dx indicating that it is "with respect to x". In integration the variable of integration is shown by adding d(variable) after the function to be integrated. Thus 23x dx means "the integral of 3x2 with respect to x”.

There are other functions that we can differentiate to obtain 3x2, for example

3 3 35, 0,5 and 2.x x x In fact, any function of the form 3x C , where C is a

constant, will be the answer to the question find 23x dx .

To allow for the possible presence of a constant, whenever the process of integration is performed, a constant, usually a C, is added to the result. We refer to C as the constant of integration. Therefore the correct mathematical notation is 2 33x dx x C .

This reverse process, where we need to add a constant of integration, is called indefinite integration. So if we write ( )g x dx ,

denotes that we have to find an indefinite integral, g(x) is the integrand, dx indicates the variable with respect to which the integrand is integrated, in this case x. Now ( )g x dx is read as: the integral of the function g(x) with respect to x

or the integral g(x) dee ex.



139

ACTIVITY 1 1a) State the derivative of x2. b) Find the indefinite integral of 2x. Write your answer in the correct mathematical notation. 2a) State the derivative of 1

3 cos3x .

b) Find the indefinite integral of sin 3x. Write your answer in the correct mathematical notation. 3. Consider the indefinite integral 23 2 3t t dt

a) State the independent variable. b) State the integrand. Remember to check the response on page 141.

2. THE GENERAL SOLUTION OF INTEGRALS OF THE FORM nax

The general solution of integrals of the form nax dx , where a and n are constants is given by

1

1

nn ax

ax dx Cn

This rule is true when n is fractional, zero, or a positive or negative integer, with the exception of n = 1.

Example 1

a) 1 1

2

1

2

2

1 1

2

2

2 2x

x

x dx x dx C

C

x C

b) 0 0

0

0 1

3 3 Remember definition 1

Thus 1 can be written as

Use as is the independent variable

31

3

dx x dx a

x

x x

xC

x C

c) 3

23 12

3 322

44 4.

1x

xdx x dx C



140

12

1

2

12

21

1

2

4.

4.

8

xC

x C

x C

Note that the constant multiple a or the coefficient of xn may be written outside the

integral sign, that is

1

1

nn n ax

ax dx a x dx Cn

.

Thus example c) above can be written as

3

23 1

2 23

2

1

32

44 4. 8

1

xdx x dx C x C

x

Remember, each result may be checked by differentiating the answer.

ACTIVITY 2 Find the integrals and check your answer:

a) 3 dx b) 22x dx

c) 2

1dx

x

d) y dy

e) 0.2x dx f) 0.2x dy


When a sum of several terms is integrated, the result is the sum of the integrals of the separate terms. Some examples will explain this more clearly.

Example 2

a) 1

22 2 3Integrate 3 5 7 8 with respect to x x x x x x :

12

12

12

32

2 2 3

2 2 3

12 1 1 1 2 1 3 1

12

23 2 1

3 5 7 8

3 5 7 8

3 5 87

2 1 1 1 1 2 1 3 1

5 2 87

2 3 2

x x x x x dx

x dx x dx dx x dx x dx x dx

x x x x xx C

x xx x x x C



141

Note that when an integral contains more than one term, there is no need to have a

constant for each term. A single constant at the end is sufficient.

b) 2 2

3 21

3 2

3 2 1 3 2 1

3 2 13 2

x x dx x dx x dx dx

x xx C

x x x C

As you get more familiar with the process, you can eliminate some steps when

writing down your answer. Remember that another person should be able to read and comprehend your answer.

c)

2

2

Find 3.7 2.9

3.7 2.9

2.93.7

2

3.7 1.45

x dx

dx x dx

xx C

x x C

ACTIVITY 3

a) 23 2 3t t dt b) 4

22

32

5

xx dx

x



3.1 Activity 1

1a) From our knowledge of differentiation 2 2d

x xdx

b) Indefinite integration reverses the process of differentiation, so we can write 22x dx x C

We always include the additional constant of integration when finding indefinite integrals.

2a) From our knowledge of differentiation

1 13 3cos3 3 ( sin 3 ) sin 3

dx x x

dx

b) Indefinite integration reverses the process of differentiation, so we can write 13sin 3 cos3x dx x C

3a) The independent variable is t. We can tell this by inspecting the term dt. b) The integrand is 23 2 3t t .



142

3.2 Activity 2

a) 03 3dx x dx

0

0

0 1

3 1 Remember 3 is a constant

Constants can be taken out of the integral

3 The independent variable is

We therefore write 1 as

30 1

3

Check: 3 3

dx

x dx x

x

xC

x C

dx C

dx

b) 2 1

2

3

22

2 1

2

3

xx dx C

xC

3 2

22 2.3Check: 2

3 3

d x xC x

dx

c) 22

1Use rules for exponentsdx x dx

x

-2 1

1

1 2 22

=2 1

11

1 1Check: 1. 1. 1

xC

xC

Cx

d dx x x

dx x dx x

d) 12y dy y dy

12

32

32

1

1 12

32

2

3

yC

yC

yC

12

32

32

1

2Check:

3

3 2.

2 3

d yC

dx

y

y

y



143

e) 0.2 1

0.2

1.2

0.2 1

1.2

xx dx C

xC

1.2

1.2 1

0.2

Check:1.2

1.21.2

d xC

dx

x

x

f) 0.2 0.2

0,2

1 The independent variable is ,

therefore must be considered as a constant

x dy x dy y

x

0.2 0

0 10.2

0.2

0.2 0.2

0 1

Check:

x y dy

yx C

x y C

dx y C x

dy

3.3 Activity 3

a) 2

3 2

3 2 3

3

t t dt

t t t C

3 2

2

Check: 3

3 2 3

dt t t C

dt

t t

b) 4

22

32

5

xx dx

x

3 5 2 1

3 5 1

3 5

2 1. 3

3 5 5 2 1

2 3

3 25 1

2 3

3 25

x x xC

x x xC

x xC

x

3 5

2 42

42

2

2 3Check:

3 25

2.3. 5( 1)3

3 25

32

5

d x xC

dx x

x xx

xx

x

You have finished learning unit 1 if you are able to explain what is meant by an indefinite integral explain what is meant by the term "integrand" explain what is meant by a constant of integration find solutions to integrals of the form nax dx

We further investigate the reverse of differentiation in learning unit 2.


INTEGRATION Reverse of

differentiation II

CONTENTS PAGE

1. SIMPLIFY THE INTEGRAND BEFORE INTEGRATION ............................. 145 2. DETERMINE THE VALUE OF THE CONSTANT OF INTEGRATION ....... 147 3. APPLICATION: DISPLACEMENT, VELOCITY, ACCELERATION ............ 148 4. RESPONSES TO ACTIVITIES .......................................................................... 149 4.1 Activity 1 ............................................................................................................. 149 4.2 Activity 2 ............................................................................................................. 150 4.3 Activity 3 ............................................................................................................. 150

MODULE 7

LEARNING UNIT 2

OUTCOMES

At the end of this learning unit, you should be able to simplify an integrand to the form axn to find a given integral find the value of the constant of integration, if more information about the integral is

known find the integral equivalents of velocity and displacement

Module 7 Learning unit 2 INTEGRATION: Reverse of differentiation II


145

1. SIMPLIFY THE INTEGRAND BEFORE INTEGRATION There is no simple rule for the integration of a product or quotient as in differentiation. In many problems we have to rely on simplifying the integrand to the general form axn.

Example 1

a) 22Find 1x x dx

The integrand is not in the general form axn and looks complicated. We know how

to expand 221 x and then multiply the answer by 12x x . So let’s apply our

knowledge and hope that we find an integral that we know how to solve.

1

2

5 912 2 2

22 2 4

22 2 4

Simplify the integrand:

1 1 2

1 1 2

2

All the above terms are of the form n

x x x

x x x x x

x x x

ax

Now we are able to find the integrand using the general solution from integration in unit 1.

5 912 2 2

3 7 112 2 2

3 7 112 2 2

22

3 7 112 2 2

1 2

2

2 4 2

3 7 11

x x x x x dx

x x xC

x x xC

3 7 112 2 2

5 912 2 2

5 912 2 2

2 4 2Check:

3 7 11

2 3 4 7 2 11. . .

3 2 7 2 11 2

2

d x x xC

dx

x x x

x x x

Note that this check only confirms the correct value of the integral

5 912 2 22x x x dx and not the simplification steps. We have to be very careful

when simplifying, as we usually cannot check that part of the answer easily.



146

b) 3 2

6

3 2 8 1Find

x x xdx

x

Start by splitting the integrand into four fractions. Divide each term of the

numerator by the denominator.

3 2 3 2

6 6 6 6 6

3 4 5 6

3 4 5 6

3 23 4 5 6

6

2 3 4 5

3 2 8 1 3 2 8 1That is

Now simplify each fraction

3 2 8 1

Use rules for exponents

3 2 8

3 2 8 1Thus 3 2 8

3 2 8

2 3 4 5

x x x x x x

x x x x x

x x x x

x x x x

x x xx x x x dx

x

x x x x

2 3 4 53 2 12

2 3 5

C

x x x x C

c) 2

32

3

1 7Find

3

x xdx

x

We can simplify the integrand under the integral sign.

23

23

5 823 3 3

5 823 3 3

7 4 13 3 3

7 4 13 3 3

73

2

3

2

3

3

3 3 3

1

73

1 7 Thus

3

1 14 49

3

14 49

3

1 14 49

3 3 3

14 49

3 3 3

14 49

3 3 3

14

3 1

x xdx

x

x x xdx

x

x x xdx

x

x x x dx

x x xdx

x x xdx dx dx

x x

4 13 3

4 23 3

13

1 1

4 13 3

49

3 1 3 1

4914

4 2

xC

x xx C



147


a) 31 u du

b) 1 2x x x dx

c) 3 26 5x x x

dxx


2. DETERMINE THE VALUE OF THE CONSTANT OF INTEGRATION To determine the value of the constant term, more information must be given in the question.

Example 2

2If 3 2 1 , find the function if 10 when 1u x x dx u u x .

We are given the value of the integral for a specific value of the independent variable. With this extra information we will be able to find the value of the constant of integration.

2 3 2

3 2

3 2

3 2 1

Thus (*)

Given 10 if 1

Substitute in *

10 1 1 1

now can be determined

7

Answer: 7

x x dx x x x C

u x x x C

u x

C

C

C

u x x x

ACTIVITY 2 1. 2If 3 10 , find the function if 0 when 2y x x dx y y x

2. 3 2If 4 3 6 2 , determine the value of when 4,

given that when 2, 30

y x x x dx y x

x y




148

3. APPLICATION: DISPLACEMENT, VELOCITY, ACCELERATION In the previous units on differentiation you saw that there are several important formulae connecting acceleration, velocity and displacement. With the knowledge you have gained so far you can write integral equivalents.

The velocity v of an object is the derivative of its displacement s, that is ds

vdt

.

The displacement s is therefore given by s v dt .

The acceleration a is the derivative of the velocity v, that is dv

adt

.

The velocity v is therefore given by v a dt .

Example 3 A ball is projected along a frictionless inclined plane. The velocity of the ball is given by

24v t t in metres per second. What is the change in displacement in the first 4 seconds?

2

2 3

As , we have .

Thus 4

4

2 3

dsv s v dt

dt

s t t dt

t tC

To find the change in displacement we need the displacement at time t = 0 and t = 4. 2 3

0

2 3

4

4 0

4(0) (0)At 0,.

2 30 0

4(4) (4)At 4,.

2 3192 128

664

6Now the change in displacement

64

6

10,6 metres

t s C

C

C

t s C

C

C

s s s

C C



149

ACTIVITY 3

1. An object is moving in a straight line with a velocity 1

m/sv tt

.

a) If the object began at the origin, what is its position when t = 15? b) What is its acceleration when t = 15? 2. An object moving with velocity u at time t = 0 accelerates with constant

acceleration a. The velocity of the object at a later time t is v. a) Use indefinite integration to find an expression for v. b) Given that when t = 0 the velocity is u, find the constant of integration. Remember to check the response on page 150.


4.1 Activity 1

a) Simplify the integrand. The binomial theorem should be used, or write

3

3 2 3

3 2 3

42 3

1 1 1 1

1 1 3 3

All the above terms are of the form

1 1 3 3

3

2 4

n

u u u u

u u u u

ax

u u u u du

uu u u C

b) 1 2x x x dx

2

3 2

4 3 2

43 2

3 2

3 2

3 2

4 3 2

4

x x x dx

x x x dx

x x xC

xx x C

c) 3 2 3 26 5 6 5x x x x x x

dx dxx x x x

2

32

6 5

3 53

x x dx

xx x C

4.2 Activity 2

1. 2

2 33 10 102

xx x dx x x C



150

23

23

23

Thus 10 (*)2

Substitute in * given 0 when 2

20 2 10 2

20 8 2 20

now can be determined

10

Answer: 10 102

xy x x C

y x

C

C

C

C

xy x x

2. 3 2 4 3 6 2 , y x x x dx

4 3 2

4 3 2

4 3 2

3 24

4 3 2

4 3

4 3 6= 2

4 3 2

3 2

given that when 2, 30 we find the value of

3 2

30 (2) 2 3 2 2 2

30 16 8 12 4

30 16

14

Thus 3 2 14

and the value of when 4

4 (4)

x x xx C

x x x x C

x y C

y x x x x C

C

C

C

y x x x x

y x

y

23(4) 2 4 14

256 64 48 8 14

246

4.3 Activity 3 1a) Position refers to displacement.

At the origin implies that at the start v = 0, t = 0 and s = 0.

12 2

12

1

2

At the origin 0, 0 thus in the above equation 0 0 0 and 0

t ts t dt C

t

s t C C



151

12 2

2

2At 15,

2 1

(15)2 15

2120.3 m

t tt s

b) Given the velocity, you have to find the acceleration, that is differentiation.

1

2

3

2

3

3

2

1

11

2

11

21

At 15, 12 (15)

0.99 m/s

dv d da t t t

dt dt dtt

t

t

t a

2. a) The velocity is found by integrating the acceleration, that is v a dt .

You are told in the question that a is constant, thus a can be taken out of the

integral.

0

1

Thus where is the constant of integration.

v a dt

a dt

a t dt

at C

v at C C

b) Evaluate the equation in a) with t = 0 and v = u.

(0)

Thus

or in the more usual form

u a C

u C

v at u

v u at

This is the end of this learning unit, so you should be able to simplify an integrand to the form axn to find a given integral find the value of the constant of integration, if more information about the integral is

known find the integral equivalents of velocity and displacement Next you will study learning unit 3 on the method of substitution.


INTEGRATION Method of substitution

CONTENTS PAGE

1. METHOD OF SUBSTITUTION ........................................................................ 153 2. RESPONSE TO ACTIVITY ............................................................................... 156

MODULE 7

LEARNING UNIT 3

OUTCOME At the end of this learning unit, you should be able to determine integrals of the form

'n

f x f x dx using the substitution u = f(x).

Module 7 Learning unit 3 INTEGRATION: Method of substitution


153

1. METHOD OF SUBSTITUTION The general solution and examples we have done up to now are useful but they do not

show us how to integrate 310 4 3 3 233 7 , 6 4 , or 4 8 .x dx x x dx x x dx The

first integral could be found by expanding 103 7x and then integrating, but this would

be a very long and inefficient method. This method could also work on the last two integrals. A better method would be to use substitution. The method of substitution involves introducing a function that changes the integrand, such that the general solution will work when integrating. The method is illustrated in the next example.

Determine 103 7x dx .

10 10

10

In this integral we will use the substitution 3 7

Differentiating with respect to we get 3

3

and 3

Now substitute in the original integral

3 7 Let 3 7 and 33

1

3

u x

dux

dxdu dx

dudx

dux dx u u x du dx

u du

10 10

Remember constants can be taken out

After substitution the new integrand should only contain constants and the new

independent variable . The new integral is of the form .

1Thus 3 7

3

nu ax

x dx u du

11

11

11

1

3 11

33Substituting for we obtain the answer in the original variable

3 7

33

uC

uC

u x

xC

Module 7 Learning Unit 3 INTEGRATION: Method of substitution


154

We substituted not only for the integrand, but also for the differential. The most difficult part is selecting the correct substitution. The substitution we select must allow us to write

as nf x dx u du with the possible exception of a multiplicative constant.

A more general form of the basic rule is

1

, 11

nn u

u du C nn

Example 1

a) Determine 34 36 4x x dx

4

3

3

3

34 3 3 4 3

In this integral we will use the substitution 6

Differentiating with respect to we get 4

4

and 4

Now substitute in the original integral

6 4 Let 6 and 4

u x

dux x

dx

du x dx

dudx

x

x x dx u du u x du x dx

4

44

4Substituting for we obtain the answer in the original variable

6

4

uC

u x

xC

b) Determine 3 23 4 8x x dx .

3

2

2

2

Let 4 8,

then 12

and 12

or 12

u x

dux

dxdu

dxx

dux dx

1

3 2 3 23 3

1

3

Thus 4 8 4 8 .

.12

x x dx x x dx

duu



155

4

3

43

4

3

4

3

43 3

433

1.

12

3

48

16

4 8

16

4 8

16

uC

uC

uC

xC

xC

c) Find 2

1

3dx

x

If we let u = x2 3, then du = 2x dx. But 2x is not a constant, so we cannot write this integrand as a constant multiple of un du. We therefore cannot use this substitution method and we cannot integrate using the methods learnt thus far.

d) Evaluate 23 5x dx .

If we let u = x3 + 5, then du = 3x2 dx. Again, the substitution method cannot be used as we cannot write this integrand as a constant multiple of un du. However, we can expand and simplify the integrand using the method in unit 2.

23 6 3

7 4

7 4

5 10 25

1025

7 4

525

7 2

x dx x x dx

x xx C

x xx C

Using the above examples, we can write a new general solution in f(x) notation as follows:

1

11

nn f(x)

f(x) .f '(x) dx c, nn

Any integral of this form can be solved by the substitution u = f(x). Note: Example c) and d) could not be solved with this method because the integrand did not contain f '(x) or a multiple of it.



156

ACTIVITY 1 Determine the following integrals:

a) 22 2 2x x dx b) 2 3 5 2 3x x x dx

c) 3 1x dx d) 32 2x x dx

e) 225 x dx f) 1

dt

t

g) 2

1

60 100 9dx

x x

h) 3 1 2x dx

i) 21x x dx j) 71 x dx

k) 2

3

3 6

2 12 10

xdx

x x

l) 3

1 44 1 x dx

m) 3

2 21x x dx n)

2

23

3

1

xdx

x

o) 42 315 1x x dx p) 22 3(1 3 ) 4x x x dx

q) 21 2

dx

x

2. RESPONSE TO ACTIVITY

a) 22 2Let 2 2 and let 2I x x dx u x

2

3

32

Then 2

and 2

Thus our integral becomes

3

2

3

dux

dxdu x dx

I u du

uC

xC

b) 2 3 5 2 3I x x x dx

2Let +3 +5

Then 2 3

u x x

du x dx

12

32

322

Therefore

2

3

2 3 5

3

I u du u du

uC

x xC



157

12

32

32

c) Let 3 1 and let 3 1

Then 3 differentiate with respect to on both sides

1

3

1

31 2

.3 32

3 19

I x dx x u

dux

dx

I u du

u du

u C

x C

32 2

32

3

4

42

d) Let 2 and let 2

Then 2

and 2

12 2

21

2

1

2 4

2

8

I x x dx u x

dux

dxdu x dx

I x x dx

u du

uC

xC

e) 2 2Let 25 and 25I x dx u x

then 2

2

dux

dxdu x dx

But 2x is not a constant, so we cannot write this integrand as a constant multiple of un du. We therefore cannot use this substitution method and we cannot integrate using the methods learnt thus far.

f) Let and 1 , then 1

dtI t u dt du

t

12

12

12

12

Thus

2

2 1

duI u du

u

u C

t C



158

g) 2 2

1Rearrange the denominator

60 100 9 9 60 100

dxdx

x x x x

2Factorise the denominator

3 10

dx

x

2

213

11 13 1

1

1Let and 3 10

60 100 9then 3

Thus

.

3 10

3

I dx u xx x

du dx

I u du

u C

xC

h) 3 1 2x dx

12

13 1 3

2

4

31 32 4

43 38

43 38

Let 1 2 , then 2 or

Thus 1 2

1 2

1 2

u x du dx du dx

x dx u du

u C

x C

x C

i) 21x x dx

2 12Let 1 , then 2 or u x du x dx du x dx

12 1 2

2

31 2 22 3

321 2

3

3213

Thus 1

1

1

x x dx u du

u C

x C

x C

j) 71 x dx

77

818

818

Let 1 , then

Thus 1

1

u x du dx

x dx u du

u C

x C



159

k) 2

3

3 6

2 12 10

xdx

x x

3 2

212

121 1 22 23

1

1 2 22 1

13 2

3

Let 2 12 10, then (6 12)

thus (3 6)

3 6Thus

2 12 10

2 12 10

2 12 10

u x x du x dx

du x dx

x dudx u du

ux x

u C

x x C

x x C

l) 3

1 44 1 x dx

331 1 444 4

11 4 44 1

1

4

Let 1 , then

Thus 1

1

u x du dx

x dx u du

u C

x C

m) 3

2 21x x dx

2 12

332 1 22

2

5

1 2 22 5

52 2

Let 1 , then 2 or

Thus 1

1

5

u x du x dx du xdx

x x dx u du

u C

xC

n)

2

23

3

1

xdx

x

3 2

22

23

1

3

Let 1 , then 3

3Thus

1

( 1)

1

1

u x du x dx

xdx u du

x

u C

Cx



160

o) 42 315 1x x dx

3 2

4 42 3 2 3

4

515

53

Let 1 , then 3

Thus 15 1 5 3 1

5

5

1

u x du x dx

x x dx x x dx

u du

u C

x C

p) 22 3(1 3 ) 4x x x dx

3 2

22 3 2

1

13

3

Let 4 , then (1 3 )

Thus (1 3 ) 4

1

4

1

(4 )

u x x du x dx

x x x dx u du

u C

x x C

Cx x

q) 21 2

dx

x

12

2122

112

Let 1 2 , then 2 or

Thus 1 2

( 1)

1

2 1 2

u x du dx du dx

dxu du

x

u C

Cx

This the end of learning unit 3 and you should therefore be able to determine integrals of

the form 'n

f x f x dx using the substitution u = f(x).

Now we move on to learning unit 4 with a focus on standard integrals.


INTEGRATION Standard integrals

CONTENTS PAGE

1. TABLE OF STANDARD INTEGRALS ............................................................ 162


f (x)dx

f(x) .......................................................... 163

3. INTEGRALS OF THE FORM f xf ' x e dx .................................................. 165

4. INTEGRALS OF THE FORM f(x)f (x).a dx .................................................. 166

5. RESPONSES TO ACTIVITIES .......................................................................... 168 5.1 Activity 1 ............................................................................................................. 168 5.2 Activity 2 ............................................................................................................. 168 5.3 Activity 3 ............................................................................................................. 169 5.4 Activity 4 ............................................................................................................. 169

MODULE 7

LEARNING UNIT 4

OUTCOMES

At the end of this learning unit, you should be able to use a table of standard integrals to solve integrals simplify an integrand to reduce to a standard integral use the method of substitution to reduce an integrand to a standard integral

Module 7 Learning unit 4 INTEGRATION: Standard integrals


162

1. TABLE OF STANDARD INTEGRALS Every derivative, when written in reverse, gives us an integral. To assist with the process of integration we resort to tables of standard integrals. These tables are generated by using a list of standard derivatives. The following list of standard integrals will be available in your examination.

TABLE OF INTEGRALS

1

1

1 11

2 11

3

4

5

6 sin cos

7 cos sin ( )

8

(n )n

nn

f(x) f(x)

f(x)f(x)

a x. ax dx c, n

n

f(x). f(x) .f'(x) dx c, n

n

f (x). dx n f(x) c

f(x)

. f (x).e dx e c

a. f (x).a dx c

n a

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f x c

. f (x)

2

2

tan sec cos

9 cot sin

10 sec sec tan

11 cosec osec cot

12 sec tan

13 cosec

. f(x) dx n f(x) c n f(x) c




. f (x). f(x) dx f(x) c

. f (x). f(x) dx

cot

14 sec tan sec

15 cosec cot cosec

f(x) c

. f (x). f(x). f(x) dx f(x) c

. f (x). f(x). f(x)dx f(x) c



163

It is important to note that when dealing with the trigonometric functions, the variable x must be measured in radians and not degrees. The modulus (absolute value) sign is inserted to remind us that only logarithms of positive numbers can be found. When using the table of standard integrals we will use the substitution u = f(x). After you have practised a lot, you will find that the substitution becomes unnecessary. You will be able to multiply by the correct constant to find f ' (x) and read off the answer to the integral straight from the table. We have already discussed integrals of standard forms 1 and 2. In this unit we will concentrate on integrals of forms 3 and 4.


f (x)dx

f(x)

In standard forms 1 and 2 we excluded the case n = . The reason is that it does not yield valid results. For n = division by 0 is obtained, which is not defined.

Remember: 11 1If , then thus

dyy n x dx x dx n x

dx x x

. We can therefore

see number 3 in the table as the special case where n = .

In general, '( ) '( ) thus ( )

( ) ( )

d f x f xn f x dx n f x

dx f x f x

.

Example 1

a) Determine 2

1 2

7

xdx

x x

2

2

2

Put 7 , then 1 2 or 1 2

1 2 1Thus

7

7

duu x x x du x dx

dxx

dx duux x

n u C

n x x C

Some of you might recognise immediately that 2the derivative of 7 is 1 2 .x x x

2

'( )Therefore the integrand is of the form

( )

and the answer can be read from the table as

7

f x

f x

n x x C



164

b) Determine

1

1dx

x x

This is more difficult to recognise, so we need to do a substitution.

11 22

11 22

1

2

Put 1 , then

and

1.

2

2

1 1.

1 1

12

12

'Now it is possible to recognise that the integrand is of the form

2

2 1

duu x x

dx

du x dx

dx

xdx

dux

dxdx

xx x x

duu

duu

f x

f x

n u C

n x C

ACTIVITY 1 Determine a) 1u du

b) 3

4

3

1

xdx

x

c) 3 2

2

3 3 1x x xdx

x




165

3. INTEGRALS OF THE FORM f xf ' x e dx

If , then from which it follows that

In general the standard integral is obtained as follows:

If , then ' , thus '

x x x x

f x f x f x f x

dyy e e e dx e C

dx

dyy e f x e f x e dx e C

dx

Example 2 a) 3Find .xe dx

The most direct route would be to say, let 3 then ' 3.

We need to modify the integrand by multiplying by the correct constants.

Remember we are using an equal sign, so we need to observe equalit

f x x f x

3 3 31 13 3

3 313

y.

(3) 3

The integrand is of the form ' .

Use the table of standard integrals to read off the answer.

.

x x x

f x

x x

e dx e dx e dx

f x e

e dx e C

13

3

13

13

313

Or we may use substitution.

Let 3 , then 3

3 or

1Hence, .

3x u

u

u

x

duu x

dxdu dx

dx du

e dx e du

e du

e C

e C

b) 2

Find .2xe x dx .

2 2

2If , then ' 2 , the integrand is of the form '

Use the table of standard integrals to write down the answer.

.2

f x

x x

f x x f x x f x e

e x dx e C



166

2

2

2

Or use substitution:

Let , then 2

Hence .2x u

u

x

u x du x dx

e x dx e du

e C

e C

c) Determine 3

1

7 xdx

e

.

We will find the answer without substitution.

3173

31 17 3

3121

3121

1

7

3

3

xx

x

x

x

dx e dxe

e dx

e dx

e C

ACTIVITY 2 Determine a) 7 5 9,2 23 11 5x x x xe e e e dx

b) 21

1 2x x

e x dx


4. INTEGRALS OF THE FORM f(x)f (x).a dx

If , then from which it follows that .

Sometimes we would slightly modify the expression to give a neater standard integral.

In this case is a constant, which can there

x x x xdyy a a n a a n a dx a C

dx

n a

fore be taken out of the integral

Dividing both sides by :

where the last term can be replaced with a single constant symbol.

In general the standard integral is

x x

xx

f

n a a dx a C

n a

a Ca dx

n a n a

f (x).a

f(x)(x) a

dx Cn a



167

Example 3 3

13

3 13

13

3

a) Find

Let 3 , then 3 or

.

3

x

x u

u

x

a dx

duu x dx du

dx

a du a du

aC

n a

aC

n a

b) Determine 23 1.7 xx dx .

2

2

2

3 1 16

16

3 1

duPut 3 1, then 6 and

6

Thus .7 7

7

7

7

6 7

x u

u

x

duu x x x dx

dx

x dx du

Cn

Cn

ACTIVITY 3 Determine a) 2xa dx

b) 53 x dx

Remember to check the response on page 169. To consolidate what you have learnt thus far, try an activity with mixed questions.


a) 2 2 127 1x xxe xa x x dx

b) 31x xe e dx

c) 1

2

xedx

x

d) 2

3

2 1x xdx

x

e) 3

4

4

4

xdx

x

f) 2

1 2

1

xdx

x x

g) 1 x x dx h) 23 4s ds

i) 23 1 2x x dx




168


5.1 Activity 1

a) 1 1u du du nu C

u

b) 3

4

4 3 3

3

4

34

434

3Determine

1

Put 1 , then 4 , thus 4

Substitute values then

3 13

41

3 1

4

1

xdx

x

du duu x x x dx

dx

x dudx

ux

duu

nu C

n x C

c)

3 2 3 2

2 2 2 2 2

2

21

2

First simplify the integrand:

3 3 1 3 3 1

33

You should be able to do this without any substitution

3 32

13 3

2

x x x x x xdx dx

x x x x x

x x dxx

xx n x x C

xx n x C

x

5.2 Activity 2 a) 7 5 9,2 2

7 5 9,2 2

75 9,2 2

3 11 5

3 11 5

7 5 9, 2 2

3 11 5

7 5 9,2 2

x x x x

x x x x

xx x x

e e e e dx

e e e eC

ee e e C



169

b)

2

2

2

1

2

1

1

Determine 1 2

Put 1 , then 1 2 and 1 2

Thus 1 2

x x

x x u

u

x x

e x dx

duu x x x du x dx

dx

e x dx e du

e C

e C

5.3 Activity 3 a) 2 21

2

2

2

2

x x

x

a dx a dx

aC

n a

b) 5 515

5

3 5.3

3

5 3

x x

x

dx dx

Cn

5.4 Activity 4

a) 2 2 12

2

7 1

Put , then 2 , and 2

x xxe xa x x dx

du duu x x xdx

dx

2 2

2

2

12

171 12 2 2

71 12 2 2

2712 2

Thus 7 1

(1 )

1

12

x x

u u

uu

xx

xe xa x x dx

e du a du u du

ae n u C

n a

ae n x C

n a

b) 31x xe e dx

Put 1,

then

and

x

x

x

u e

due

dx

du e dx



170

3 3

4

4

Thus 1

4

1

4

x x

x

e e dx u du

uC

eC

c) 1

2

xedx

x

1

1

1 2 22

2

Put , then 1 and

Thus x

u

u

x

du dxu x x du x dx

dx x

edx e du

x

e C

e C

d) 2

3 2 3

2 1 1 2 1x xdx dx

xx x x

2 3

1 2

2

12

2

1 22 1

2

x x dxx

x xn x C

n x Cx x

e) 3

4

4

4

xdx

x

4

'The integrand is of the form

4

f x

f x

n x C

f) 2

1 2

1

xdx

x x

2

'The integrand is of the form

1

f x

f x

n x x C

g) 1 x x dx



171

1 3

2 2

3 5

2 2

[using the law . ]

2 2

3 5

m n m n

x dx x x dx

x dx x dx a a a

x x C

h) 2 2133 4 3 3 4s ds s ds

3

13

3

2 2

3 2

3 2

3 3

3 4

3

3 4

9or

3 4 9 24 16

9 2416

3 2

3 12 16

The two answers should be the same.

To compare expand the fraction in answer one using the binomial theorem:

3 4 3 3(3

9

sC

sC

s ds s s ds

s ss C

s s s C

s s s

3(2)2 2 32

3 2

3 2 649

) (4) (3 )(4) 4

9

27 108 144 64

9

3 12 16 , where is incorporated in

sC

s s sC

s s s C C

i) 1

2 2 23 1 2 3 1 2x x dx x x dx

123 2

4

323 2 2

4 3

3212

4 1 2

1 2

1 2

x x dx

x

x C

Since this is the end of learning unit 4, you should be able to use a table of standard integrals to solve integrals simplify an integrand to reduce to a standard integral use the method of substitution to reduce an integrand to a standard integral

Learning unit 5 deals with partial fractions.


INTEGRATION Partial fractions

CONTENTS PAGE

1. INTEGRATION USING PARTIAL FRACTIONS ............................................ 173 2. RESPONSE TO ACTIVITY ............................................................................... 175

MODULE 7

LEARNING UNIT 5

OUTCOME At the end of this learning unit, you should be able to use partial fractions to integrate algebraic expressions.

Module 7 Learning unit 5 INTEGRATION: Partial fractions


173

1. INTEGRATION USING PARTIAL FRACTIONS The first step is to verify that you do not have an integral of one of the following forms: EXAMPLE

1. ' , 1n

f x f x dx n

5252

42

2 22 2 2

2

2

4

xdx x x x dx

x x

x xC

'

2.f x

dxf x

2

2

2 22

2

xdx n x x C

x x

Partial fractions may be used in integration of algebraic expressions as follows:

Example 1

a) Find the integral of 2

3 8

2 35

xdx

x x

with the aid of partial fractions.

The integral is not of the forms mentioned above. We can return to the examples in module 3, unit 2 to see how to write the expression in partial fractions.

2

3 8 13 23 That is,

12 7 12 52 35

x

x xx x

This gives us two integrals of form 2.

2

13 2312 12

3 8 13 23

12 7 12 52 35

7 5

xdx dx dx

x xx x

n x n x C

b) Use partial fractions to find

2

4

3 7

1

xdx

x

The integral is not of the forms mentioned above. We can return to the examples in module 3, unit 2 to see how to write the expression in partial fractions.

2

4 2 3 4

3 7 3 6 4

1 1 1 1

x

x x x x

This gives us three integrals of form 1.



174

2

4 2 3 4

2 3

2 3

3 7 3 6 4

1 1 1 1

3 6 4

1 2 1 3 1

3 3 4

1 1 3 1

xdx dx dx dx

x x x x

Cx x x

Cx x x

c) Find 3 2

2

2 4 4

2

x x xdx

x x

.

The integral is not of the forms mentioned above. The integrand is an improper fraction. We can return to module 3, unit 3 to see how to write the expression in partial fractions.

3 2

2

2 4 4 4 33

2 12

x x xx

x xx x

This gives four integrals

3 2

2

2

24 3

42

3

2 4 4 4 33

2 12

3 4 2 3 12

using the rules of logarithms we can combine

the two terms with logarithms

3 2 ( 1)2

23

2 1

x x xdx x dx dx dx dx

x xx x

xx n x n x C

xx n x n x C

xxx n C

x

ACTIVITY 1 Find

2

3 2

2

2

33

2

2 8a)

5 6

3 7 4b)

2

8c)

2

4 8d)

4 5

x

x x

x x xdx

x x

xdx

x

xdx

x x




175

2. RESPONSE TO ACTIVITY a) First resolve the integrand in partial fractions:

2

2 8 2 8

3 2 3 25 6

Combine right-hand side:

2 32 8

3 2 3 2

Equating numerators gives:

2 8 2 3

Let 2, then 2(-2) 8 ( 2 3)

4

Let 3, then 2( 3) 8 ( 3 2)

2

2

2 8

3

x x A B

x x x xx x

A x B xx

x x x x

x A x B x

x B

B

x A

A

A

x

x x

2

2 4

2 3 2

The integral becomes

2 8 2 4

3 25 6

2 43 2

2 3 4 2

using the rules of logarithms we can simplify our answer

by combining the two logarithm ter

x x

xdx

x xx x

dx dx

x x

n x n x C

2 4

4 2

4

2

ms

3 2

2 3

2

( 3)

n x n x C

n x n x C

xn C

x



176

b) Resolve the integrand in partial fractions. The degree of the numerator is larger

than the degree of the denominator. It is an improper fraction. Therefore we first

divide:

2 3 2

3 2

2

2

1

2 3 7 4

2

7

2

5 4

x

x x x x x

x x

x x

x x

x

3 2

2 2

2

2

3 7 4 5 4Thus 1

2 2

5 4Resolve in partial fractions

25 4 5 4

2 22

Combine right-hand side:

25 4

2 2

Equating numerators gives:

5 4 2

Let 2, then 5 2 4

x x x xx

x x x x

x

x xx x A B

x x x xx x

A x Bxx

x x x x

x A x Bx

x

2

2

2 6

3

Let 0, then 4 2

2

5 4 2 3Therefore

22

B

B

B

x A

A

x

x xx x



177

3 2

2

2

The integral becomes

3 7 4 2 31

22

1 12 3

2

2 3 22

using the rules of logarithms we can simplify our answer

by combining the two logarith

x x xx dx

x xx x

x dx dx dx dxx x

xx n x n x C

2

32

m terms

22

xx n x x C

c) This is an example of standard form 2 with n a negative number.

232 38

333

23

23

83 2

2

28

3 2

4

3 2

xdx x x dx

x

xC

Cx

d) This is an example of standard form 3.

2 2

2

4 8 2 42

4 5 4 5

2 4 5

x xdx dx

x x x x

n x x C

This is the end of learning unit 5, so you should be able to use partial fractions to integrate algebraic expressions in the form of a quotient. Learning unit 6 deals with trigonometric functions.


INTEGRATION Trigonometric

functions

CONTENTS PAGE

1. INTEGRATION OF TRIGONOMETRIC FUNCTIONS .................................. 179 2. RESPONSES TO ACTIVITIES .......................................................................... 182 2.1 Activity 1 ............................................................................................................. 182 2.2 Activity 2 ............................................................................................................. 184

MODULE 7

LEARNING UNIT 6

OUTCOMES

At the end of this learning unit, you should be able to integrate trigonometric functions using a table of standard integrals use trigonometric identities to rewrite the integrand in standard form use the method of substitution to simplify the integrand

Module 7 Learning unit 6 INTEGRATION: Trigonometric functions


179

1. INTEGRATION OF TRIGONOMETRIC FUNCTIONS In this unit we will discuss the basic trigonometric integrals. The following list of standard integrals will be available in your examination. Number 6 to 15 deal with trigonometric functions. The substitution u = angle will usually allow an easier integration.

TABLE OF INTEGRALS

1

1

1 11

2 11

3

4

5

6 sin cos

7 cos sin ( )

8

(n )n

nn

f(x) f(x)

f(x)f(x)

a x. ax dx c, n

n

f(x). f(x) .f'(x) dx c, n

n

f (x). dx n f(x) c

f(x)

. f (x).e dx e c

a. f (x).a dx c

n a

. f (x). f(x) dx f(x) c

. f (x). f(x) dx f x c

. f (x)

2

2

tan sec cos

9 cot sin

10 sec sec tan

11 cosec osec cot

12 sec tan

13 cosec

. f(x) dx n f(x) c n f(x) c




. f (x). f(x) dx f(x) c

. f (x). f(x) dx

cot

14 sec tan sec

15 cosec cot cosec

f(x) c

. f (x). f(x). f(x) dx f(x) c

. f (x). f(x). f(x)dx f(x) c



180

Example 1 a) Find sin 5x dx

Let 5 , then 5 and 5

1Hence sin 5 sin .

51

cos51

cos55

du duu x dx

dx

x dx u du

u C

x C

This is an example of standard form 6. b) Find cos 2 1x dx

Let 2 1, then 2 and 2

1Hence cos 2 1 cos .

21

sin21

sin (2 1)2

du duu x dx

dx

x dx u du

u C

x C

This is an example of standard form 7. c) 2Find sin tan secx x dx

2

2

Let tan , then sec

sin tan sec sin

cos

cos tan

u x du x dx

x x dx u du

u C

x C

ACTIVITY 1 Find a) cos 7x dx

b) tan 5 7x dx

c) 2cotx x dx

d) sec 3 4 tan 3 4x x dx

e) 2 2 3secx x dx

f) 2 2cosecx x dx




181

Example 2 a) tan 2Find .secxe x

2

tan 2

tan

Let tan , then sec

.secx u

u

x

u x du x dx

e x dx e du

e C

e C

b) sin cos

Find cos

x xdx

x

First simplify the integrand

sin cos sin cos

cos cos cos

tan 1

tan 1

We can now use the table of standard integrals.

= sec

x x x xdx dx

x x x

x dx

x dx dx

n x x C

b) Evaluate 2tan 2 1x dx .

2 2

2

2

2

First expand the integrand as we cannot recognise a standard form.

tan 2 1 tan 2 2 tan 2 1

tan 2 1 2 tan 2

sec 2 2 tan 2

sec 2 2 tan 2

Now let 2 , then 2 and 2

Subs

x dx x x dx

x x dx

x x dx

x dx x dx

duu x du dx dx

2

titute these values in the above integrals:

1sec tan

21

tan sec21

tan 2 sec 22

u du u du

u n u C

x n x C



182

ACTIVITY 2 Determine a) sin cosxa x dx

b) 3cos 2 sin 2xe x dx

c) cosx xe e dx

d) 2tan x dx

e) 2

1

cosdx

x

f) 2cos sinx x dx

g) 4sec 3 tan 3x x dx

h) 2sin x dx

i) 2tan 3 sec 3x x dx

j) 1 cos

sin

xdx

x



2.1 Activity 1 a)

Find cos 7

Let 7 , then 7 and 7

1Hence cos 7 cos .

71

sin71

sin 77

x dx

du duu x dx

dx

x dx u du

u C

x C

b)

Find tan 5 7


x dx

du duu x dx

dx



183

1Hence tan 5 7 tan .

5This compares with standard form 8.

1sec

51

sec 5 75

x dx u du

n u C

n x C

c)

2

2

2 12

212

Find cot

Let , then 2

cot cot

This is an example of standard form 9.

sin

x x dx

duu x x dx

x x dx u du

n x C

d)

13

13

13

Find sec 3 4 tan 3 4


sec 3 4 tan 3 4 sec tan

This is an example of standard form 14.

sec

sec 3 4

x x dx

duu x du dx dx

x x dx u u du

u C

x C

e)

2 2 3

3 2 2

2 2 3 2

3

Evaluate sec

Let , then 3 and 3

1Therefore sec sec

3This is an example of standard form 12.

1tan

31

tan3

x x dx

dux u du x dx x dx

x x dx u du

u C

x C



184

f)

2 2

2

2 2 2

2

Find cosec

Let , then 2 and 2

1Therefore cosec cosec

2This is an example of standard form 13.

1cot

21

cot2

x x dx

duu x du x dx x dx

x x dx u du

u C

x C

2.2 Activity 2 a) sinFind cosxa x dx

sin

sin

Let sin , then cos

Hence cosx u

u

x

u x du x dx

a x dx a du

aC

n a

aC

n a

b) 3cos 2Find sin 2xe x dx

3cos 2

3cos 2

3cos 2

Let 3cos 2 , then 6sin 2 and sin 26

1Hence sin 2

61

6

6

x u

x

x

duu x du x dx xdx

e x dx e du

e C

eC

c)

Find cos

Let , then

Hence cos cos

sin

sin

x x

x x

x x

x

e e dx

u e du e dx

e e dx u du

u C

e C



185

d)

2

2 2

2 2

2

Find tan

This does not seem to fit in one of the standard forms.

But, if we remember that tan sec 1,

we see that

tan sec 1

sec 1

Now we have standard form 12 in the first integr

x dx

x x

x dx x dx

x dx dx

al

= tan x x C

e)

2

2

1sec

costan

dx x dxx

x C

f)

2

2 2

3

3

Find cos sin

Let cos , then sin and sin

Therefore cos sin

3

cos

3

x x dx

u x du x dx du xdx

x x dx u du

uC

xC

g)

4

4 3

Find sec 3 tan 3

This is difficult to recognise.

Remember that the derivative of sec sec tan

Rewrite the integrand sec 3 tan 3 sec 3 sec3 tan 3

Let sec3 , then 3sec3 tan 3 and sec33

x x dx

x x x

x x dx x x x dx

duu x du x x dx x

4 3

3

4

4

tan 3

Therefore sec 3 tan 3 sec 3 sec3 tan 3

1

3

12

sec 3

12

xdx

x x dx x x x dx

u du

uC

xC



186

h)

2

12 2

2

2

Find sin

Let , then and 2

Therefore sin 2 sin

2cos

2cos

x

x

x

x

dx

u du dx du dx

dx u du

u C

C

i)

2

2

2

2

Find tan 3 sec 3

If tan 3 , then ' 3sec 3

Therefore tan 3 sec 3

1tan 3 3sec 3

3This is an example of standard form 8

1sec3

3

x x dx

f x x f x x

x x dx

x x dx

n x C

j)

1 cos 1 cos

sin sin sin

cosec cot

cosec cot sin

We can simplify our answer using logarithm rules

= sin cosec cot

x xdx dx dx

x x x

x dx xdx

n x x n x C

n x x x C

You have now reached the end of learning unit 6, so you should be able to integrate trigonometric functions using a table of standard integrals use trigonometric identities to rewrite the integrand in standard form use the method of substitution to simplify the integrand

The next learning unit will focus on the definite integral.


INTEGRATION The definite integral

CONTENTS PAGE

1. DEFINITION ...................................................................................................... 188 2. EVALUATION OF A DEFINITE INTEGRAL ................................................. 188 3. RESPONSE TO ACTIVITY ............................................................................... 190 4. EXERCISE 1 ....................................................................................................... 191

MODULE 7

LEARNING UNIT 7

OUTCOMES

At the end of this learning unit, you should be able to define a definite integral explain the meaning of the terms “upper limit” and “lower limit” calculate the value of a definite integral

Module 7 Learning unit 7 INTEGRATION: The definite integral


188

1. DEFINITION You are already familiar with the procedure to evaluate an indefinite integral:

If ' then .g x f x f x dx g x C

Now we can define the definite integral of f(x) between the limits x = a and x = b as:

b

a

b

a

f x dx g b g a

g x

The constants a and b are called the integration limits. A is called the lower limit and b is called the upper limit. The integration constant C does not play a role in the calculation of a definite integral. The integration constant appears in both brackets and disappears when simplified:

b

af x dx g b C g a C

g b C g a C

g b g a

If the limits of a definite integral are switched around, the sign of the answer changes,

that is b a

a bf x dx f x dx .

2. EVALUATION OF A DEFINITE INTEGRAL (i) Find the indefinite integral (omitting the constant of integration) and enclose

within square brackets with the limits at the right-hand end, or put a straight line with the limits at the right-hand end.

or b b

a af x f x

(ii) Substitute the upper limit b for x in f(x), that is find f(b). (iii) Substitute the lower limit a for x in f(x), that is find f(a). (iv) Subtract f(a) from f(b).

Example 1

a) Calculate 3

14 2x dx

2

3 32

11

2 2

Since 4 2 2 2

it follows from the definition that

4 2 2 2

2 3 2 3 2 1 2 1

12 0

12

x dx x x C

x dx x x



189

b) Determine 1

34 2x dx

2

1 12

33

2 2

Since 4 2 2 2


4 2 2 2

2 1 2 1 2 3 2 3

0 12

12

x dx x x C

x dx x x

Note: Examples a) and b) illustrate that b a

a bf x dx f x dx . Furthermore, the

definite integral in example b) is a negative number. There are no restrictions on

the value of a definite integral. In example b), the lower limit 3 is larger than the

upper limit 1. This is permissible. The choice of upper and lower limits depends

on the application that is being made of the definite integral.

c) Find 2

0cos 2x dx

22

12

12 00

1 122 2

1 12 2

Since cos 2 sin 2


cos 2 sin 2

sin 2 sin 2(0)

sin sin 0

0 0

0

x dx x C

x dx x

d) Evaluate 2

0sin 3cos3x x dx

2

2

2

0

00

2 2

Since sin 3cos3 cos sin 3


sin 3cos3 cos sin 3

cos sin 3 cos 0 sin 3 0

0 ( 1) 1 0

1 1

1 1

2

x x dx x x C

x x dx x x



190

ACTIVITY 1 Calculate

a) 2 2

1 3 x x dx

b) 0 22

11x x dx

c) 2

20

sin 2

1 cos

xdx

x

d) 1

2 2

-14 xe dx

3. RESPONSE TO ACTIVITY

a) 23 2

2 2

11

33 =

3 2

x xx x dx

3 2 3 22 3 2 1 3 1

3 2 3 2

8 1 36

3 3 2

16 36 2 9

641

65

66

b) 0 02 22 2

1 1

11 2 1

2x x dx x x dx

032

1

032

1

3 32 2

11

2 3

11

61

0 1 1 161

1 86

7

6

x

x



191

c) 22

2 20 0

2sin .cossin 2

1 cos 1 cos

x xxdx dx

x x

22

01 cos

1 0 1 1

1 2

2

12

0,693

n x

n n

n n

n

n

d) 1 1

2 22 2

-1 1

4 14 2

1 2x xe dx e dx

122

1

1 2

2

2

2

12

5.166

xe

e e

ee

4. EXERCISE 1

Evaluate each of the given definite integrals:

2

34

2

2 12 2

1 1

2 03 22 23

1 1

13

2 3 23

10

6

a) 3 b) 2

c) 4 3 d) 1

1 1e) f)

g) h) sin2

x

x dx x x dx

x x dx x x dx

dx e dxx x

dxx dx

x

ANSWERS

1 2 1a) 5 b) 4 c) 7.5 d) 0.3 e) 1

3 3 9f) 4.99 g) 0.693 h) 0.707



192

Learning unit 7 is now complete and you should be able to define a definite integral explain the meaning of the terms “upper limit” and “lower limit” calculate the value of a definite integral

Learning unit 8 is next, and you will learn how to calculate the areas using integration.


INTEGRATION Areas

CONTENTS PAGE

1. INTRODUCTION ............................................................................................... 194 2. SUMMATION AND THE DEFINITE INTEGRAL .......................................... 195 3. THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE ................. 196 4. RESPONSES TO ACTIVITIES .......................................................................... 205 4.1 Activity 1 ............................................................................................................. 205 4.2 Activity 2 ............................................................................................................. 206 5. EXERCISE 1 ....................................................................................................... 208

MODULE 7

LEARNING UNIT 8

OUTCOMES

At the end of this learning unit, you should be able to interpret a definite integral geometrically find the area under a curve find the area between curves

Module 7 Learning unit 8 INTEGRATION: Areas


194

1. INTRODUCTION When a graph is used to represent a physical quantity, it is often useful to calculate the area under the graph, because this can represent a related physical quantity. For example, the area under a graph of velocity against time represents the distance covered. Consider the following graph:

To find the distance travelled, the area under the graph can be calculated by dividing the area into geometric shapes:

The area of rectangles and triangles can be found easily. Now consider:

Velocity

Time

Velocity

Time

Velocity

Time

v(t)



195

The area now depends on the curve v(t) and it is no longer possible to calculate the area exactly and thus the distance covered using simple geometric shapes. In unit 2 of this

module we calculated distance by integrating the velocity, recall that s v dt . This

suggests a relationship between the area under a graph and integration.

2. SUMMATION AND THE DEFINITE INTEGRAL Consider the area bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b, where b > a. (See figure 1.)

Figure 1 Divide the interval into n equal parts each of length x.

Figure 2 Since the areas of the strips in figure 2 are unknown, we propose to approximate each strip by a rectangle whose area can be found.

a bX

Y

0

f(x)

a bX

Y

0x



196

In figure 3 a representative strip and its approximating rectangle are shown.

Figure 3 In figure 3, suppose the representative strip is the i th strip counting from the left, and let x = xi be the coordinate of the midpoint of its base. Let yi = f(xi) be the ordinate of the

point Pi (on the curve) whose coordinate is xi. Draw a line through Pi parallel to the x-

axis and complete the rectangle MRSN. The area of the i th strip can now be approximated by the area of rectangle MRSN. Area of the th strip = height of strip length of strip

i

i

i

y x

y x

Note that the error in the calculated area and the actual area under the curve will become smaller if we decrease the width of the strip x. When each strip is treated similarly, it seems reasonable to take the sum of all the strips as an approximation of the total area under the curve between x = a and x = b We can write an approximation of the required area as

1 2 31

.....n

n ii

y x y x y x y x y x

[The sigma sign is used to abbreviate the sum.] Now suppose that the number of strips (with approximating rectangles) is indefinitely increased so that the width of the strips becomes infinitely small, that is x 0 if n . It is evident from the figure that by so increasing the number of approximating rectangles, the sum of their areas more nearly approximates the required area, that is

a bX

Y

0xM N

SR

x

Pi (xi;yi)

xi



197

1

Area limn

in

i

b

a

b

a

y x

y dx

f x dx

Thus the summation process is a special integration process. In fact, the elongated s- symbol is meant to indicate a type of summation.

3. THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE

Geometrically, b

af x dx is interpreted as the area between the curve f(x), the x-axis

and the points a and b on the x-axis. The value of the integral is positive if the area is above the x-axis and it is negative if the area is below the x-axis. Therefore we need to take the absolute value when calculating areas using a definite integral.

If 0, then 0 :b

af x f x dx

If 0, then 0 and area = :b b

a af x f x dx f x dx

a bX

Y

0

f( x)

a bX

Y

0

f( x)



198

The steps necessary for setting up the definite integral which yield a required area are as follows: (i) Make a rough sketch showing

the required area a representative strip the approximating rectangle, with reference point P(x; y)

(ii) Write the area of the approximating rectangle. (iii) Assume the number of rectangles to be indefinitely increased and evaluate the

definite integral.

Example 1: Area between a straight line and the axes We will use this example to show that we get the same answer using the formula for the area of a triangle, summing over vertical strips and summing over horizontal strips. Find the area between the straight line with x-intercept = 4 and y-intercept = 3 and the x- and y-axes. We draw a sketch of the required area:

We can find the equation of the straight line AB using ordinary geometry as 3

34

y x

Method 1: Using the formula

12

2

Area of width height

4 3

2

6 units

AOB



199

Method 2 Using vertical strips: Sketch the required area and a representative strip.

Area of representative strip = height width

33

4

y x

x x

Therefore the required area A can be calculated with the definite integral:

4

0

4

0

42

0

2

2

33

4

33

8

3 43 4 0 0

8

6 12

6 units

x

xA y dx

x dx

xx



200

Method 3: Using horizontal strips Sketch the required area and a representative strip.

Area of representative strip A = x . y

Required area 3

0

y

yA x dy

Note: (i) The limits MUST be in terms of y because we integrate with

respect to y. (ii) We MUST express x in terms of y. We change the subject of the equation of the straight line to x.

33

43

34

44

3

xy

xy

yx

Therefore the required area is

3

0

3

0

32

0

32

0

32

0

2

44

3

44

3 2

24

3

2 34 3

3

6 12

6 units

y

yA x dy

ydy

yy

yy



201

Example 2: Calculating distance from a velocity time graph

The velocity v of a body t seconds after a certain instant is ( 22 5t ) m/s. Find by integration how far it moves in the interval from t = 0 to t = 4 s.

Since 22 5t is a quadratic expression, the curve v = 22 5t is a parabola cutting the v-axis at v = 5. The distance travelled is given by the area under the v/t curve.

4 4 2

0 0

43

0

33

Thus distance = 2 5

25

3

2 02(4)5(4) 5 0

3 3

12820 0

3

62.67 m

v dt t dt

tt

Remember to use the correct unit. As you become more familiar with calculating areas, you can omit showing the representative strip. All the necessary information is contained in the definite integral.



202

Example 3: Area between a curve and the axes

Find the area bounded by the curve 2y x and the x-axis, between x = 1 and x = 2.

We first sketch the curve and the area to be calculated, as well as a representative strip.

The required area is shown (shaded) above.

Area of representative strip A = x .y

2

1

2 2

1

23

1

3 3

Required area

3

2 1

3 38 1

3 37

31

2 square units3

x

xA y dx

x dx

x

y = x2



203

ACTIVITY 1 Find the area bounded by the curve siny x and the x-axis between 0x

and x .


Example 4: Area between curve and x-axis where the curve lies partly above and below the axes

Consider now a curve which lies partly above and partly below the x-axis. At some values of x, the y-ordinate of the curve will be negative. If we evaluate the area of a strip at these values of x, we will clearly get a negative result since x is always taken as positive and this positive length multiplied by a negative length will yield a negative product. In cases like these, the "negative" areas and positive areas must be evaluated separately and the absolute value of the two added to give the total area. This then explains the absolute necessity for a suitable sketch. Find the area bounded by the curve y = sin x and the x-axis between x = 0 and x = 2. We first sketch the curve, paying particular attention to the values where the curve

crosses the x-axis.

2

0

2

0

2

0

Area of representative strip =

.

sin sin

cos cos because cos cos

cos cos0 cos 2 cos

1 1 1 1

2 2

4 square units

x x

x x

y dx

Area y dx y dx

x dx x dx

x x x x



204

+The necessity for a sketch and the above method of evaluation is immediately evident if

we evaluate 2 2

0 0sin

x

xy dx x dx

.

2 2

0 0

2

0

sin

cos

cos 2 cos 0

1 1

0

x

xy dx x dx

x

An area of 0 is obviously not true.

ACTIVITY 2 1. Find the area between the curve y = cos x and the x-axis between x = 0 and x = 2. Find the area between the curve 2 2 3f x x x and the x-axis

between 2x and x = 4. Remember to check the response on page 206.

Example 5: Area between two curves Calculate the area enclosed by the parabolas 2 26 and 2y x x y x x . Sketch the parabolas on the same set of axes.

Find the intersection of the parabolas by setting the equations equal:

2 2

2

2

2 6

2 8 0

4 0

4 0

0 or 4

x x x x

x x

x x

x x

x x

Find the y-values by substituting into one of the equations.



205

Thus the intersections of the parabolas are at (0;0) and (4;8).

Length of a typical strip: 2 21 2 6 2k k k ky y x x x x

Required area 4

1 20y y dx

4 2

0

42 323 0

23

2

8 2

4

4(16) 64

64

3

21.33 units

x x dx

x x


4.1 Activity 1

Sketch the required area.

0

0

0

Area of representative strip : .

Required area

sin

cos

cos cos0

1 1

2 square units

x

x

A y x

A y dx

x dx

x



206

4.2 Activity 2

1. We first sketch the area.

2

2

2

2

1 1 0

0

2 2

1 2

0

: cos

sin

1 square unit

: cos

sin

1square unit

Total area

1 1

2 square units

As before: cos 0 which is the algebraic sum of the two areas.

A A x dx

x

A A x dx

x

A A

x dx

In the above case, it was not necessary to evaluate both areas since one is obviously equal to the other by virtue of the symmetry of the cosine curve. When we are dealing with symmetric curves it is sometimes convenient to evaluate the area under a specific portion of the curve and then to multiply this by a constant. 2. To sketch the area you must first recognise the curve as a parabola. Therefore we

need to find the x- and y-intercepts and the turning point.



207

2

2

-intercepts Put 0 : 2 3 0

3 1 0

1 or 3

-intercept Put 0 : 3

Turning point Use differentiation or the formulae from school

' 2 2 and 1 1 2 1 3

2 2 0 4

1

Coordinates of turning poin

x f x x x

x x

x

y x y

f x x f

x

x

t 1; 4 .

1

2

1 2

2

132

2

813 3

2

1

2 3

33

1 3 4 6

2 21

3 3

12 units

3

x

xArea y dx

x x dx

xx x



208

3

1

3 2

1

332

1

27 13 3

4

3

4 2

3

432

3

64 273 3

2

2 3

33

9 9 1 3

29 1

3

210

3

210

3

3

2 3

33

16 12 9 9

26 9

3

12

3

x

x

x

x

Area y dx

x x dx

xx x

Area y dx

x x dx

xx x

Area Ar

1 2 3

115 square units

3

ea Area Area

5. EXERCISE 1

a) Find the area enclosed between the x-axis and 2 4y x .

b) Find the area between the x-axis and 3y x from x = 1 to x = 1

c) Find the area bounded by the parabola 24 x y and the y-axis using (i) a vertical representative strip (that is perpendicular to the y-axis) (ii) a horizontal representative strip (that is perpendicular to the x-axis)

d) Find the area between the curve 23 2 y x x and the x-axis.

e) Calculate the area between the curve 3 2

3 3 1xy x x , the x-axis and the lines

x = 1 and x = 3.



209

ANSWERS

2

2

2

2

a) 10.67 square units

1b) units

22

c) 10 units3

d) 10,67 units

e) 12 units

Now that you have come to the end of this learning unit, you should be able to interpret a definite integral geometrically find the area under a curve find the area between curves

Now do the post-test below.


POST-TEST: INTEGRATION 1. Determine

2

2a

2

(b) 1 1

3 4 7c

xx dx

x

x x dx

x xdx

x

2. Determine z if 3 29 11 3 , given that when 1, 2.z y y y dy y z

3. When an object having initial velocity u accelerates with constant acceleration a,

the velocity of the object at time t is s. The velocity is the derivative of

displacement, that is ds

vdt

(a) Derive an expression for s. (b) Given that when 0t the displacement 0s , find the value of the

constant of integration.

4. An object moves with variable acceleration a given by the formula 23a t . (a) Find an expression for the velocity of the object. (b) Find an expression for the displacement of the object.

5. Find

22

3

2

2

a b 41

c d 19 172 3

x dxx x dx

x

dxx x dx

x

(e) 2

3

2 3

23 7

3

x

xx

(f) 323 . 3t t dt

(g) 21x xe e dx

6. Determine the following integrals, using partial fractions where necessary to

simplify the integrand:

(a)

2

2

2

1

x xdx

x

(b) 2

1

2 4

xdx

x x

(c)3 2

2

5 4x xdx

x

(d) 2

3 3

9

( 2)

xdx

x

Module 7 INTEGRATION: Post-test


211

(e) 2

4 3 2

xdx

x

(f) 2

11 3

2 3

xdx

x x

(g)

22 9 35

1 2 3

x xdx

x x x

7. Evaluate

(a) 2 42x x dx (b) 21 x

dxx

8. Determine (a) tan 2x dx (b) 23 sin 3 .cos3x x dx

(c) sin5 cosxe x dx (d) tan 2 2secxe x dx

(e) cosec

sin

xdx

x

(f) sec3 tan3

5 2sec3

x xdx

x

9. Evaluate the following integrals:

(a) 3 32

12x x dx

(b) 2 2

11x x dx

(c) 2

cos sinxe x dx

10. Find the area bounded by the parabola x = 8 + 2y y2, the y-axis and the lines y = 1 and y = 3.

11. Find the area between the curve y = x3 6x2 + 8x and the x-axis.

12. Calculate the area between the curve y = 2x2 x and the lines x =3 and x = 2.

13. Roughly sketch the curve y = 4x2 3x 1 between x = 1 and x = 2. Determine the area enclosed by the curve, the x-axis and the lines x = 1

and x = 2.


INTEGRATION: POST-TEST SOLUTIONS

1. 2

a2

xx dx

x

1 12 2

3 12 22

3 12 2

3 2

22

1. 2.

2 2

24

3 4

xx x dx

x x xC

x xx C

32

32

2

32

2

2 3

(b) 1 1

Simplify integrand before integration

= 2 1

2

2

4

2 3As the question was written in surd (root) form, we write the answer as

4

2 3

x x dx

x x dx

x xx C

x xx C

x xx C

111 122 2

1122

32

2

2

2

5 3 1

2 2 2

5 3 12 2 2

5 3 1

2 2 2

5 3

3 4 7c

3 4 7

3 4 7

3 4 7

3 4 7

6 8 14

5 3 1

6 814

5 3

x xdx

x

x xdx

x x x

x x x dx

x x x dx

x x xC

x x xC

x xx C

2. 3 29 11 3z y y y dy

Module 7 INTEGRATION: Post-test solutions


213

4 3 2

4 3 2

9 113

4 3 2 given 1, 2

9 11 12 3

4 3 29 11 1

2 34 3 2

24 27 44 6 36

125

12Thus the answer is

9 11 53

4 3 2 12

y y yy C

y z

C

C

y y yz y

3.

(a) Given that , and that , we can writeds

v u at vdt

212

212

212

212

so that

where is a constant of integration.

(b) We are given that when 0, 0, and substituting these values gives

0 0 0

0

Therefore

dsu at

dt

s u at dt

s ut at C

C

t s

s ut at C

u a C

C

s ut at

4. 2Given that 3 .a t

2 3

3 414

(a) 3

(b)

where and are constants of integration

v a dt t dt t C

s v dt t C dt t Ct D

C D

Module 7INTEGRATION: Post-test solutions


214

5.

2 2

3 3

1 3a

31 1

x dx xdx

x x

12

12

3

12

3

11.

3

21

3

xC

x C

32

32

2 2

2

32

2

1b 4 2 4

2

41

2

4

3

x x dx x x dx

xC

xC

2

2

c2 3

Let 2 3 then 6 .

This integral cannot be solved with the methods we discussed thus far;

You will learn suitable methods in Mathematics II.

dx

x

u x du x

(d) 219 17x x dx

2

12 1 2

34

3

1 2 234 3

32 2

32

Let 19 17

then, 34

and 19 17

19 17

51

19 17

51

u x

du x dx

x x dx u du

u C

xC

xC

(e)2

3

2 3

23 7

3

x

xx

322

Let 3 7 then (2 3)3

xu x du x dx .



215

122

3

1

2

13 2

3

2 3and

23 7

3

2

22 3 7

3

22 3 7

3

xu du

xx

u C

xx C

xx C

(f) 33

2 23 223 . 3 2 . 3t t dt t t dt

12 2

2

3 2. 3

2 13

3

t C

Ct

(g) 21x xe e dx

2 2

3

3

Let 1, then

Therefore 1

3

1

3

x x

x x

x

u e du e dx

e e dx u du

uC

eC

6.

(a)

2 2

2 2

2 2

2 11

x x x xdx dx

x xx

2

2

2

2

We can use partial fractions or manipulate the integrand as follows:

2 1 1

2 1

( 2 1) 1

2 1

x xdx

x x

x xdx

x x

2

2 2

2

( 2 1) 1

2 1 2 1

11

1

x xdx dx

x x x x

dx dxx



216

2

1

1

1

11

1

x x dx

xx C

x Cx

(b) 2

1

2 4

xdx

x x

121 2

2

12 2

(2 2) 2 4

1 2. 2 4

2 1

x x x dx

x x C

(c) 3 2

2

5 4x xdx

x

2

2

5 4

45

2

x x dx

xx C

x

(d) 2

3 3

9

( 2)

xdx

x

32 3

23

23

3 3 2

23.

23

2 2

x x dx

xC

Cx

(e) 2

4 3 2

xdx

x

12 3 4

33 4

334

13 2

3

1 4. 2

3 34

29

x x dx

x

x C



217

(f) 2

11 3

2 3

xdx

x x

2

2

Resolve the integrand in partial fractions

3 111 3 11 3

1 3 1 3 1 32 3

Equating numerators gives

11 3 3 1

Let 1: 8 4

2

Let 3 : 20 4

5

11 3 2Hence

12 3

A x B xx x A B

x x x x x xx x

x A x B x

x A

A

x B

B

xdx dx

xx x

2

5

5

3

2 1 5 3

1by the laws of logarithms

3

dxx

n x n x C

xn C

x

(g)

22 9 35

1 2 3

x xdx

x x x

2

2

Resolve the integrand in partial fractions

2 9 35

1 2 3 1 2 3

2 3 1 3 1 2

1 3 2

Equating numerators gives

2 9 35 2 3 1 3 1 2

Let 1: 24 6

4

Let 2 : 45 15

x x A B C

x x x x x x

A x x B x x C x x

x x x

x x A x x B x x C x x

x A

A

x B

3

Let 3 :10 10

1

B

x C

C



218

2

4

3

2 9 35 4 3 1Hence

1 2 3 1 2 3

4 1 3 2 3

1 3by the laws of logarithms

2

x xdx dx dx dx

x x x x x x

n x n x n x C

x xn C

x

7.

(a) 2 42x x dx

2 2

2 2

12 2

12 2

32 2

32

1 2

1 2

1 2

14 1 2

4

1 2. 1 2

4 31

1 26

x x dx

x x dx

x x dx

x x dx

x C

x C

(b) 2 2

1

2

1 1 2x x xdx dx

xx

1 1 3

2 2 2

1 3 54 22 2 23 5

2

2

x x x dx

x x x C

8. (a) 1 1

2 2tan 2 2 tan 2 sec 2x dx x dx n x C

(b) sin sin

If sin , then ' cos

Therefore 5 cos 5x x

f x x f x x

e x dx e C

(c)3

2 sin 33 sin 3 cos3

3

xx x dx C

(d) tan 2 tansecx xe x dx e C



219

(e) 2coseccosec cot

sin

xdx x dx x C

x

(f) sec3 tan3

5 2sec3

x xdx

x

Let 5 2sec3 , then 6sec3 tan 3

sec3 tan 3 1 1Therefore

5 2sec3 61

65 2sec3

6

u x du x x dx

x xdx du

x u

nu C

n xC

9.

(a) 3 32

1

2

Let 2

Let 2, then 2

I x x dx

duu x x

dx

2

2

The values of the limits must also be changed to correspond with the

new variable :

Therefore if 1 then 1 2 1

and if =3 then 3 2 7

u

x u

x u

7 3

1

74

1

1

2

1

8

12401 1

8300

I u du

u

(b) 2 2

1Put 1I x x dx

2Let 1, then 2

duu x x

dx .



220

12

32

3 32 2

12

2

2

2 2

1

3

0

3

32 0

1

Therefore if 1 then 1 1 0

and if =2 then 2 1 3

11.2 .

21

2

1

2

13 0

31

.3 .33

3

x u

x u

I x x dx

u du

u

(c) 2

cosPut sinxI e x dx

cos

cos

cos

Let

Then sin

and sin

x

x

x

u e

dux e

dx

du x e dx

Now substitute in the original integral.

b

aI du

where a and b are the new limits in terms of the variable u. However, in

this case we are not going to use u in our final calculation, so there is no

need to calculate the new limits.

2

cos

coscos 2

1 0

11

b

a

b

a

x

I du

u

e

e e

e e

e

210. 8 2

-intercepts: 4 2 0

4 2

y y x

y y y

y or y



221

3 2

133

2

1

13

23

Area 8 2

83

24 9 9 8 1

24 6

230

3

y y dy

yy y

3 2 211. 6 8 6 8

4 2

x x x x x x

x x x



222

2 43 2 3 2

0 2

424 43 2 3 2

0 2

Area 6 8 6 8

2 4 2 44 4

4 4

8

x x x dx x x x dx

x xx x x x

212. 2 0

2 1 0

10 of/or

2

x x

x x

x x

1 12 2

1 12 2

0 0 22 2 2

3

0 0 23 2 3 2 3 2

3

Area 2 2 2

2 2 2

3 2 3 2 3 2

1 1 1 16 1 118 4 2

2 12 8 3 2 8

1 1 8122

2 24 24

1125

12

x x dx x x dx x x dx

x x x x x x



223

2

2

13. 4 3 1

4 3 1 0

4 1 1 0

1of/or 1

4

y x x

x x

x x

x x

2 2

1

23 2

1

Area 4 3 1

4 3

3 2

32 4 36 2 1

2 3 2

2 12 1

3 6

53

6

x x dx

x xx

You have now completed study guide 2 and as such the whole module: MAT1581. Make sure that you have worked through the study guides sequentially, know the theory and are able to do the calculations as guided by the outcomes of every module.

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MATHEMATICS I (Engineering) · MATHEMATICS I (Engineering) STUDY GUIDE 2 for MAT1581 L E Greyling...

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