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
CONTENTS
PAGE
Introduction
ii
STUDY GUIDE 2
MODULE 6
MODULE 7
Differentiation
Learning unit 1
Learning unit 2
Learning unit 3
Learning unit 4
Learning unit 5
Learning unit 6
Learning unit 7
Learning unit 8
Learning unit 9
Functional notation
Limits
The derivative
Standard forms
Rules of differentiation I
Rules of differentiation II
Higher order derivatives
Applications I
Applications II: maxima and
minima
4
11
26
34
41
51
70
79
88
Post-test
109
Integration
Learning unit 1 Reverse of differentiation I
Learning unit 2 Reverse of differentiation II
Learning unit 3 Method of substitution
Learning unit 4 Standard integrals
Learning unit 5 Partial fractions
Learning unit 6 Trigonometric functions
Learning unit 7 The definite integral
Learning unit 8 Areas
Post-test
137
144
152
161
172
178
187
193
210
(i)
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
modulus of k, that is the size of k
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

limiting value as n  
lim
n

dy
dx
integral
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.
(ii)
FORMULA SHEET
ALGEBRA
Factors
Laws of indices
1.


a  b  a  b a  ab  b 
a 3  b 3  a  b  a 2  ab  b 2
a m  a n  a m n
m
2.
3.
a
 a mn
n
a
a   a
m n
3
mn
6.
7.
ab  a b
5.
a
m
and
1
a n  n
a
x  a  x  b
3
an
a
   n
b
b


2

If
ax 2  bx  c  0
then
x
If y  e x then x  n y
Laws
1.
log  A  B   log A  log B
2.
 A
log    log A  log B
B
5.
A
B
C
D



2
3
x  a  x  a  x  a  x  b 

Quadratic formula
Definitions
If y  a x then x  log a y
4.

f x 
Ax  B
C


2
ax  bx  c  x  d  ax  bx  c  x  d 
Logarithms
3.
2
f x 
A
B
C



 x  a  x  b x  c  x  a   x  b   x  c 
f x 
n
8.
2
Partial fractions
m
n
 n am
1
a n  n
a
0
a 1
4.
 
 an
3
 b  b 2  4ac
2a
log An  n log A
logb A
log a A 
logb a
a log a f  f
 e n f  f
DETERMINANTS
a11 a12 a13
a
a 21 a 22 a 23  a11 22
a 32
a 31 a 32 a 33
a 23
a
 a12 21
a 33
a 31
a 23
a
 a13 21
a 31
a 33
a 22
a 32
 a11 a 22 a 33  a 32 a 23   a12 a 21 a 33  a 31 a 23   a13 a 21 a 32  a 31 a 22 
(iii)
SERIES
Binomial theorem
a  b n  a n  na n 1b  nn  1 a n  2 b 2  nn  1n  2 a n 3 b 3  ....
2!
3!
and b  a
1  x n  1  nx  nn  1 x 2  nn  1n  2 x 3  ...
2!
3!
and  1  x  1
Maclaurin’s theorem
f 0 
f 0  2 f 0  3
f n 1 0  n 1
x 
f  x   f 0  
x
x 
x 
n  1!
1!
2!
3!
Taylor’s theorem
n 1


f a 
x  a   f a   x  a 2  f a   x  a 3    f a   x  a n1  
f  x   f a  
n  1!
1!
2!
3!
f a  h   f a  
h
h2
h n1 n1
f a  
f a    
f a   
n  1!
1!
2!
COMPLEX NUMBERS
1. z  a  bj  r cos  j sin    r   re j ,
7. De Moivre's theorem
 r   r n n  r n  cos n  j sin n 
n
where j 2  1
Modulus :
r z 
a  b 
2
2
8.
b
Argument :   arg z  arc tan
a
2. Addition :
a  jb   c  jd   a  c   j b  d 
3. Subtraction :
a  jb   c  jd   a  c   j b  d 
4. If m  jn  p  jq, then m  p and n  q
5. Multiplication : z1 z 2  r1r2 1   2 
6. Division :
1
z n has n distinct roots:
1
1
  k 360
n
with k  0, 1, 2,  , n  1
z  rn
n
9. re j  r  cos   j sin  
 
 
  re j  r cos  and  re j  r sin 
10. ea  jb  e a  cos b  j sin b 
11. n re j  n r  j
z1 r1
 1   2 
z 2 r2
(iv)
GEOMETRY
MENSURATION
1. Circle: (  in radians)
1. Straight line
y  mx  c
y  y1  m x  x1 
Perpendiculars, then m1 
Area   r 2
Circumference  2 r
1
m2
Arc length   r
1 2
1
r   r
2
2
1
Segment area  r 2   sin  
2
Sector area 
2. Angle between two lines
m  m2
tan   1
1  m1 m 2
3. Circle
2. Ellipse
x2  y2  r 2
x  h   y  k   r
2
2
4. Parabola
y  ax 2  bx  c
b
axis at x 
2a
5. Ellipse
x2 y2

1
a2 b2
6. Hyperbola
Area  ab
Circumference   a  b 
3. Cylinder
Volume  r 2 h
2
Surface area  2rh  2r 2
4. Pyramid
1
Volume  area base  height
3
5. Cone
1
Volume  r 2 h
3
Curved surface  r
6. Sphere
A  4r 2
4
V  r 3
3
7. Trapezoidal rule
xy  k
x2 y2

 1 round x - axis 
a2 b2
x2 y2
 2  2  1 round y - axis 
a
b
1 b  a 
 f  x0   2 f  x1     2 f  xn1   f  xn  
2  n  
8. Simpson’s rule
1 ba 
[ f  x0   4 f  x1   2 f  x2   4 f  x3  
3  n 
2 f  x4     2 f  xn  2   4 f  xn 1   f  xn ]
9. Prismoidal rule
b  a 
 n   f  m1   f  m2     f  mn 1   f  mn  


(v)
HYPERBOLIC FUNCTIONS
TRIGONOMETRY
Definitions
e x  ex
sinh x 
2
x
e  e x
x
cosh 
2
x
e  e x
tanh x  x
e  ex
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
tan A  tan B
tan  A  B  
1  tan A tan B
tan A  tan B
tan  A  B  
1  tan A tan B
Identities
cosh 2 x  sinh 2 x  1
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)
2 tan A
tan 2 A 
1  tan 2 A
1  tanh x  sech x
2
2
coth 2 x  1  cosech 2 x
1
sinh 2 x  cosh 2 x  1
2
1
cosh 2 x  cosh 2 x  1
2
sinh 2 x  2 sinh x cosh x
cosh 2 x  cosh 2 x  sinh 2 x
 2 cosh x  1
2
 1  2 sinh 2 x
TRIGONOMETRY
Identities
sin 2   cos 2   1
1 + tan  = sec 
2
2
cot 2  + 1 = cosec 2 
sin(-) = - sin
cos (-) =  cos
tan (-) = - tan
sin
tan 
cos
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
 x  y   x y 
sin x  sin y  2 sin 
cos

 2   2 
 x  y   x y 
sin x  sin y  2 cos
sin 

 2   2 
 x y 
x  y
cos x  cos y  2 cos
cos


 2 
 2 
 x  y   x y 
cos x  cos y  2 sin 

 sin 
 2   2 
(vi)
DIFFERENTIATION
f  x  h  f  x
dy
 lim
dx h 0
h
d
2.
k 0
dx
d n
3.
ax  anx n 1
dx
d
4.
f .g  f .g ' g . f '
dx
d f g . f ' f .g '
5.

dx g
g2
d
n
n 1
6.
 f ( x)  n  f ( x) . f '( x)
dx
dy dy du dv
7.
. .

dx du dv dx
8. Parametric equations
dy
dy
 dt
dx
dx
dt
d  dy 
2
d y dt  dx 

dx
dx 2
dt
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
1.
Test for inflection: f (b  h) and f(b + h)
Change sign or f '"(b) if f '"(b) exists
10.
d
sin 1 f ( x) 
dx
11.
d
cos 1 f ( x ) 
dx
12.
d
f '( x)
tan 1 f ( x) 
dx
1   f ( x) 2
13.
 f '( x)
d
cot 1 f ( x ) 
dx
1   f ( x) 2
14.
d
sec1 f ( x ) 
dx
15.
d
cosec1 f ( x) 
dx
16.
d
sinh 1 f ( x) 
dx
17.
d
cosh 1 f ( x) 
dx
18.
d
f '( x)
tanh 1 f ( x) 
2
dx
1   f ( x)
19.
d
f '( x)
coth 1 f ( x ) 
dx
 f ( x)2  1
20.
d
sech 1 f ( x) 
dx
21.
d
cosech 1 f ( x ) 
dx
22. Increments:
f '( x)
2
1   f ( x)
 f '( x)
2
1   f ( x) 
f '( x)
f  x
 f ( x)2  1
 f '( x)
f  x
 f ( x)2  1
f '( x)
 f ( x)2  1
f '( x)
 f ( x)2  1
 f '( x)
2
f  x  1   f ( x) 
 f '( x)
f  x
z
 f ( x)2  1
z
z
z
. x 
. y 
. w
x
 y
w
23. Rate of change:
dz  z dx  z dy  z dw

. 
. 
.
dt  x dt  y dt  w dt
INTEGRATION
b
1. By parts :  udv  uv-  vdu
3. Mean value =
2.  f(x)dx  F(b)  F(a)
a
1 b
y dx
b-a  a
4. (R.M.S.) 2 
(vii)
1 b 2
y dx
b-a  a
TABLE OF INTEGRALS
1.
 ax dx
a x(n 1 )

 c,
n 1
2.
  f(x) .f '(x) dx

3.

4.
 f '(x).e
5.

6.
 f '(x).sin f(x) dx
  cos f(x)  c
7.
 f '(x).cos f(x) dx
 sin f(x)  c
8.
 f '(x). tan f(x) dx
 n sec f(x)  c
9.
 f '(x).cot f(x) dx
 n sin f(x)  c
10.
 f '(x).sec f(x) dx
 n sec f(x)  tan f(x)  c
11.
 f '(x).cosec f(x) dx
 n  cosec f(x)  cot f(x)  c
12.
 f '(x).sec f(x) dx
 tan f(x)  c
13.
 f '(x).cosec f(x) dx
  cot f(x)  c
14.
 f '(x).sec f(x). tan f(x) dx  sec f(x)  c
15.
 f '(x).cosec f(x).cot f(x)dx  cosec f(x)  c
n
n
f ' (x)
dx
f(x)
f(x)
n  1
 f(x)n1  c, n  1
n 1
 n f(x)  c
dx
f ' (x).a f(x) dx
2
2
 e f(x)  c

a f(x)
c
n a
(viii)
M
O
D
U
L
E
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).
6
LEARNING UNIT 1 FUNCTIONAL NOTATION
1.
1.1
1.2
2.
PAGE
4
FUNCTIONS .......................................................................................................... 5
Dependent and independent variables .................................................................... 5
Functional notation ................................................................................................. 5
RESPONSES TO ACTIVITY ................................................................................ 8
LEARNING UNIT 2 LIMITS
1.
2.
3.
4.
4.1
4.2
5.
6.
6.1
6.2
6.3
11
INTRODUCTION ................................................................................................ 12
THE TANGENT QUESTION .............................................................................. 12
WHAT IS MEANT BY A LIMIT? ...................................................................... 14
TECHNIQUES FOR FINDING LIMITS ............................................................. 15
When the limit of the denominator of a quotient is 0 ........................................... 18
Limits at infinity ................................................................................................... 19
ONE-SIDED LIMITS AND CURVES ................................................................ 21
RESPONSES TO ACTIVITIES ........................................................................... 23
Activity 1 .............................................................................................................. 23
Activity 2 .............................................................................................................. 24
Activity 3 .............................................................................................................. 25
LEARNING UNIT 3 THE DERIVATIVE
1.
2.
3.
26
THE TANGENT QUESTION AND THE DERIVATIVE .................................. 27
DIFFERENTIATION FROM FIRST PRINCIPLES ........................................... 29
RESPONSES TO ACTIVITY .............................................................................. 31
LEARNING UNIT 4 STANDARD FORMS
1.
2.
2.1
2.2
1
34
STANDARD FORMS .......................................................................................... 35
RESPONSES TO ACTIVITIES ........................................................................... 39
Activity 1 .............................................................................................................. 39
Activity 2 .............................................................................................................. 39
MAT1581
Mathematics 1 (Engineering)
M
O
D
U
L
E
LEARNING UNIT 5 RULES OF DIFFERENTIATION I
1.
1.1
1.2
1.3
1.4
2.
2.1
2.2
2.3
2.4
2.5
6
DERIVATIVES OF COMBINED EXPRESSIONS ............................................ 42
Constant times a function...................................................................................... 42
Sums and differences ............................................................................................ 42
Product rule ........................................................................................................... 45
Quotient rule ......................................................................................................... 46
RESPONSES TO ACTIVITIES ........................................................................... 47
Activity 1 .............................................................................................................. 47
Activity 2 .............................................................................................................. 47
Activity 3 .............................................................................................................. 48
Activity 4 .............................................................................................................. 49
Activity 5 .............................................................................................................. 49
LEARNING UNIT 6 RULES OF DIFFERENTIATION II
1.
1.1
1.2
2.
2.1
2.2
2.3
2.4
2.5
2
70
HIGHER ORDER DERIVATIVES ..................................................................... 71
RESPONSES TO ACTIVITY .............................................................................. 77
LEARNING UNIT 8 APPLICATIONS I
1.
2.
2.1
2.2
3.
3.1
3.2
4.
4.1
4.2
4.3
51
DERIVATIVES OF COMPOSITE FUNCTIONS .............................................. 52
Function-of-a-function or chain rule ..................................................................... 52
General standard forms ......................................................................................... 61
RESPONSES TO ACTIVITIES ........................................................................... 65
Activity 1 .............................................................................................................. 65
Activity 2 .............................................................................................................. 66
Activity 3 .............................................................................................................. 66
Activity 4 .............................................................................................................. 66
Activity 5 .............................................................................................................. 68
LEARNING UNIT 7 HIGHER ORDER DERIVATIVES
1.
2.
41
79
L’HOSPITAL’S RULE ........................................................................................ 80
THE GRADIENT OF A CURVE ......................................................................... 81
The equation of a tangent to a curve ..................................................................... 81
The equation of a normal to a curve ..................................................................... 83
RATE OF CHANGE ............................................................................................ 84
Velocity ................................................................................................................. 84
Acceleration .......................................................................................................... 84
RESPONSES TO ACTIVITIES ........................................................................... 86
Activity 1 .............................................................................................................. 86
Activity 2 .............................................................................................................. 87
Activity 3 .............................................................................................................. 87
MAT1581
Mathematics 1 (Engineering)
LEARNING UNIT 9 APPLICATIONS II: MAXIMA AND MINIMA
1.
1.1
1.2
1.3
2.
3.
4.
4.1
4.2
88
MAXIMA AND MINIMA ................................................................................... 89
Definitions............................................................................................................. 89
Maximum value .................................................................................................... 90
Minimum value ..................................................................................................... 90
DERIVED CURVES ............................................................................................ 91
PRACTICAL APPLICATIONS ........................................................................... 99
RESPONSES TO ACTIVITIES ......................................................................... 103
Activity 1 ............................................................................................................ 103
Activity 2 ............................................................................................................ 107
POST-TEST
POST-TEST SOLUTIONS
MAT1581
Mathematics 1 (Engineering)
109
113
3
M
O
D
U
L
E
6
MODULE 6
DIFFERENTIATION
Functional notation
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
CONTENTS
PAGE
1.
1.1
1.2
2.
4
FUNCTIONS .......................................................................................................... 5
Dependent and independent variables .................................................................... 5
Functional notation ................................................................................................. 5
RESPONSES TO ACTIVITY ................................................................................ 8
MAT1581
Mathematics 1 (Engineering)
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 y  x3  3x 2  5 . 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 5 x 4  3x 2  4 and 3x 4  5 x3  4 x 2 y 7 we can identify them in the following
way:
f  x   5 x 4  3x 2  4 and
G  x, y  = 3x 4  5 x3  4 x 2 y 7
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 y  x3  3x 2  5 and we let f  x   x3  3x 2  5 , 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.
MAT1581
Mathematics 1 (Engineering)
5
Module 6 Learning unit 1
DIFFERENTIATION: Functional notation
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 f ( x)  x 2  1 , find the value of a) f (1)
b)
f 1  1  1  1  1  2
c)
f  3   3   1  9  1  10
d)
f  a    a   1  a2  1
b)
f (3)
and c) f (a)
2
2
2
Example 3
x2  4x  7
Given that G  x  
find G  0 , G 10 and G  x  h  .
x2
 0 2  4  0   7
G  0 
 0  2

7
 3,5
2
10   4 10   7
G 10  
10   2
2
6

100  40  7
12

147
 12, 25
12
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 1
DIFFERENTIATION: Functional notation
 x  h 2  4  x  h   7
G  x  h 
 x  h  2
x 2  2 xh  h 2  4 x  4h  7
xh2

Example 4
If angular displacement  is given by   wt  ½at 2 , 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  .
Then   f  0 
 w  0  12 a  0
2
0
Thus the displacement is 0 when the time is counted 0, which is usually at the
beginning of the movement.
Further   f 10   w 10   1 2 a 10 
2
 10 w  50a
that is, the displacement is 10 w  50a units when the time has moved 10 seconds from
the start of the counting.
ACTIVITY 1
1. Given f ( x)  5x 2  2 x  1 determine
f  0 , f  5 , f  7 , f  1 and f  x  h .
2. If   x   3x 2  5 x  8 find
a)
  1
b)
3 1
c)
  2     1
3. If F  x   6 x 3 , find
a)
3F 1  4 F  1
b)
F  x  h  F  x
c)
F h
h
MAT1581
Mathematics 1 (Engineering)
7
Module 6 Learning unit 1
DIFFERENTIATION: Functional notation
4. If g  x   2 x 2 , find expressions for
a)
g  x  h
b)
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.
c)
Calculate the value of 2  f  x    2  h  x   .
2
6. Consider the functions f (t )  t 2  4t  2 and g ( s)  s  3 .
Find the following values:
a) f (2)
b) g (9)
c) 2 f (1)  3 g (2)
7. Given that f  x   x3  3x  1 and g  t   2t  3 ,
determine the following values:
a)
f 3
b)
 
f  3
c)
g  a  1  g  a 
f  g ( 2) 
d)
Remember to check the response on page 8.
2.
RESPONSES TO ACTIVITY
Activity 1
1.
f (0)  5  0   2  0   1
2
1
f (5)  5  5   2  5   1
2
 125  10  1
 116
f (7)  5  7   2  7   1
2
 245  14  1
 232
2
f (1)  5  1  2  1  1
 5  2 1
8
8
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 1
DIFFERENTIATION: Functional notation
f ( x  h)  5  x  h   2  x  h   1
2


 5 x 2  2 xh  h 2  2 x  2h  1
 5 x 2  10 xh  5h 2  2 x  2h  1
2.
a) 10
b) 0
c) 24
3.
a) 42
b) 18 x 2h  18 xh 2  6h 3
c) 6h 2
4.
a) 2 x 2  4hx  2h 2
b) 4hx  2h 2
c) 4 x  2h
5.

  2  3x  x + 4 
 2  4 x  12 x  9 x   6 x  2 x  8
2[f (x)]2 - 2[h(x)]=2 2 x 2 + 3 x
4
2
3
3
2
3
 8 x 4  24 x3  18 x 2  6 x3  2 x  8
 8 x 4  18 x3  18 x 2  2 x  8
6.
a)
f (2) = 2² + 4(2) + 2
= 4 + 8 +2
= 14
g (9) = 9 + 3.
b)
= 12
c)
2 f (1) + 3g (2)
= 2[(1)² + 4(1) + 2] + 3[2 + 3]
= 2[7] + 3[5]
= 29
7.
a)
f
 3    3 3  3  3   1
 3 3  3 3 1
1
MAT1581
Mathematics 1 (Engineering)
9
Module 6 Learning unit 1
DIFFERENTIATION: Functional notation
b)

 

3


f  3   3  3  3 1
 3 3  3 3  1
c)
g  a  1
1
g a
  2  a  1  3   2a  3
 2a  2  3  2 a  3
d)
2
First calculate g  2 
g  2 
 2  2   3
 4  3
 7
Thus f  g ( 2)   f  7 
  7   3  7   1
3
321
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.
10
MAT1581
Mathematics 1 (Engineering)
MODULE 6
LEARNING UNIT 2
DIFFERENTIATION
Limits
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
CONTENTS
PAGE
1.
2.
3.
4.
4.1
4.2
5.
6.
6.1
6.2
6.3
11
INTRODUCTION ................................................................................................ 12
THE TANGENT QUESTION .............................................................................. 12
WHAT IS MEANT BY A LIMIT? ...................................................................... 14
TECHNIQUES FOR FINDING LIMITS ............................................................. 15
When the limit of the denominator of a quotient is 0 ........................................... 18
Limits at infinity ................................................................................................... 19
ONE-SIDED LIMITS AND CURVES ................................................................ 21
RESPONSES TO ACTIVITIES ........................................................................... 23
Activity 1 .............................................................................................................. 23
Activity 2 .............................................................................................................. 24
Activity 3 .............................................................................................................. 25
MAT1581
Mathematics 1 (Engineering)
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
 x1 , y1  and  x2 , y2  are two points on a line, then the slope of the line is given by
m
y2  y1
.
x2  x1
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 y  2 x  2 , which is a straight line.)
12
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
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
MAT1581
Mathematics 1 (Engineering)
13
Module 6 Learning unit 2
DIFFERENTIATION: Limits
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 f  x   2x  1 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
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
14
x>2
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
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
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 lim  2 x  1  3 .
x2
In general we write lim f  x   L to say that the limit of f (x), as x approaches a, is L.
xa
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 C  C , where a and C are real numbers
x a
lim x  a, where a is a real number
x a
Examples: lim 7  7 and lim x  3
x2
x 3
In the following rules we will assume that
lim f  x   L1 and lim g  x   L2 where L1 and L2 are real numbers.
xa
xa
MAT1581
Mathematics 1 (Engineering)
15
Module 6 Learning unit 2
DIFFERENTIATION: Limits
Rule 3: Limit of a sum or difference
lim  f  x   g  x    lim f  x   lim g  x 
x a
x a
x a
 L1  L2
Thus to find the sum (or difference) of two functions, you can find the sum (or
difference) of their limits.
Example: lim  x  2   lim x  lim 2
x 3
x 3
x 3
 rule 3
 rule 1 and 2
 3 2
5
Rule 4: Limit of a product
lim  f  x  .g  x    lim f  x  .lim g  x 
x a
x a
x a
 L1 .L2
This rule states that the limit of the product of two functions is the product of their
limits.
Example: lim 3x   lim 3   lim x 
x  1
 x  1   x  1 
  3 1
 rule 4
 rule 1 and 2
 3
Rule 5: Limit of a quotient
f  x L
f  x  lim
 xa
 1 ,if L2  0
lim
x a g  x 
lim g  x  L2
x a
If the limit of the denominator is not 0, then the limit of the quotient of two functions
is the quotient of their limits.
lim  2 x  4 
2x  4
 x 2
Example: lim
 rule 5
x2 x  1
lim  x  1
x2

2(2)  4
(2)  1
 rule 1 to 4
Rule 6: Limit of  f  x   or n f  x 
n
If n is a positive integer
16
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
n
n
n
lim  f  x     lim f  x     L1 
xa
 xa

lim n f  x   n lim f  x   n L1
xa
x a

Examples: lim x 3  lim x
x 5
x 5
   5 and
3
3
lim 3 4 x  3 lim 4 x  3 8  2
x 2
x 2
Example 1
a)

lim 2 x 3  5 x
x 3
  lim 2 x  lim 5x
3
x 3
x 3
 2  3  5  3 
3
 39


b) lim x 3  4 x 2  6  lim x 3  lim 4 x 2  lim 6
x 3
x 3
x 3
x 3
  3  4  3  6
3
2
 3
c)


x2  1
x 2  1 lim
x 2
lim

x 2 4 x  1
lim  4 x  1
x 2
lim x 2  lim1
x 2
 x 2
lim 4 x  lim1
x 2

(2)2  1
(4)2  1

5
7
ACTIVITY 1
Determine
a) lim y 2  y
y 2

x 2

b) lim  x  3  x  4 
x 3
c) lim 3k 2
k 5
2x2  x  5
d) lim
x 1
x4  3
Remember to check the response on page 23.
MAT1581
Mathematics 1 (Engineering)
17
Module 6 Learning unit 2
DIFFERENTIATION: Limits
4.1
When the limit of the denominator of a quotient is 0
2  x  4
2
To find the limit of lim
we find that we cannot use rule 5. Checking the
x
denominator we see that lim x  0 . But this does not mean that the limit does not
x 0
x 0
exist. If we simplify the fraction, we find that
 2  x 2  4
x

4  4 x  x2  4
x

4 x  x2
x

x 4  x
x
 4 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.
2
2  x  4
Now lim
 lim  4  x   4
x 0
x 0
x
Thus whenever substitution results in 00 , we must do more work to determine whether
the limit exists.
Example 2
a)
 x  2  x  2 
x2  4
 lim
x 2 x  2
x 2
x2
lim
Note that the denominator becomes 0 and starts to simplify the fraction.
The factor  x  2 can be cancelled.
This factor is sometimes called the vanishing factor.
 x  2  x  2 
x2  4
 lim
Thus lim
x 2 x  2
x 2
x2
 lim  x  2 
x 2
 22
4
b)
x3
x3
 lim
2
x 3 x  9
x 3  x  3 x  3
lim
1
x 3 x  3
 lim

18
1
6
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
c)
x 1
 lim
x 1 x  1
x 1
lim
x 1
 x  1 x  1
 lim
x 1

1
11

1
2
Treating x  1 as the difference between squares
1
x 1
ACTIVITY 2
Determine
3 x 2  x  10
a) lim 2
x  2 x  5 x  14
x 2  8 x  12
x  2 3 x 2  x  10
b) lim
x 2  3x  2
x 0
x3  1
c) lim
Remember to check the response on page 24.
4.2
Limits at infinity
n
n
1
1
1
1

, n  0 . Now lim n  lim     lim 
x  x n
x  x
x   x 
x

x

1
Remember that y  is the standard form of a rectangular hyperbola. Using the
x
1
graph of the hyperbola we find that lim  0 .
x  x
n
n
1
1
n
1

Thus lim n  lim     lim    0   0 .
x  x
x   x 
 x  x 
1 1 1
Notice from this example that as x  , 2 , 3 , 4 ,... all have limits of 0.
x x x
n
c
1

Using this with rule 3, we see that lim n  lim c  lim   c.0  0 .
x  x
x 
 x  x 
Consider lim


Thus we have the following which we will use to find limits at infinity:
Limits at infinity
c
0
x  x n
If c is a constant then lim
MAT1581
Mathematics 1 (Engineering)
19
Module 6 Learning unit 2
DIFFERENTIATION: Limits
Example 3
a)

lim x 2  3 x  2
x 
 Take out the highest power of x as a common factor.
  3 2 
 lim  x 2  1   2  
x 
x x 
 
    1  0  0 
2

b)
x 2  3x  2
x 
x3  1
We will first divide both the numerator and denominator by the largest power
of x in the denominator, in this case x3 . 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.
x 2 3x 2
 3 3
3
x 2  3x  2
x
x
x
lim
 lim
3
3
x 
x

1
x 1
x
 3
3
x
x
lim
 lim
x 
1
x
3
 2 x3
x2
1  1 x3
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
x 2  3x  2 0  0  0
lim

x 
1 0
x3  1

0
1
0
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
3x3  4 x 2  x  1
a) lim
x 
2 x3  2 x  1
b) lim 4  5 x 2  x 3  4 x 4
x 


Remember to check the response on page 25.
20
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
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 lim f  x   1 .
x 0
On the other hand, as x approaches 0 from the left, f (x) approaches –1 and we write
lim f  x   1 .
x 0
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
lim f  x  does not exist.
x 0
As a second example of a one-sided limit, consider f  x   x  3 as x approaches 3
(see figure below). Since f is defined only when x  3, we speak of the limit as x
approaches 3 from the right. From the diagram it is clear that lim x  3  0.
x 3
Figure 7
MAT1581
Mathematics 1 (Engineering)
21
Module 6 Learning unit 2
DIFFERENTIATION: Limits
A third example: Refer to figure 8.
1
Now let's look at y  f  x   2 near x  0. Figure 4 below shows the graph of the function.
x
Notice that as x  0, both from the left and from the right, f (x) increases without bound.
Hence no limit exists at 0. We say that as x  0, f (x) becomes positively infinite
1
and symbolically we write lim 2  
x 0 x
Figure 8
A fourth example as shown in figure 9: the graph of the hyperbola.
1
for x  0.
x
1
As x approaches 0 from the right, becomes positively infinite.
x
Consider the graph of y  f  x  
Figure 9
22
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
1
becomes negatively infinite.
x
1
1
Symbolically we write lim   and lim  .
x 0 x
x 0 x
As x approached 0 from the left,
1
does not exist.
x 0 x
Either one of these facts implies that lim
6.
RESPONSES TO ACTIVITIES
6.1
Activity 1
a)
lim y 2  y  lim y 2  lim y
y 2


y 2
y 2
 22  2
6
b)
lim  x  3 x  4   lim  x  3 .lim  x  4 
x 3
x 3
x 3
  lim x  lim 3 . lim x  lim 4 
x 3   x 3
x 3 
 x 3
  3  3 3  4 
 6
c)
lim 3k  3lim k
2
k 5
2
k 5
 3  25 
 75
2x  x  5
2 x 2  x  5 lim
x 1
lim

x 1
x4  3
lim x 4  3
2
d)
x 1

2
2
1
MAT1581
Mathematics 1 (Engineering)
23
Module 6 Learning unit 2
DIFFERENTIATION: Limits
6.2 Activity 2
a)
3x 2  x  10
x  2 x 2  5 x  14
lim
 3x  5 x  2 
x  2  x  7  x  2 
 lim
3x  5
x 2 x  7
 lim
b)
x 2  8 x  12
x  2 3 x 2  x  10
lim

3 2  5
27

11
9
 x  2  x  6 
x  2  x  2  3 x  5 
 lim
x6
x 2 3 x  5
 lim

26
3(2)  5

c)
x 2  3x  2
lim
x 0
x3  1
4
11
2
0  3 0  2


 0 3  1

2
1
2
Note in this case we can use rule 5, as the denominator does not become 0.
24
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 2
DIFFERENTIATION: Limits
6.3
Activity 3
a)
3x3  4 x 2  x  1
lim
x 
2 x3  2 x  1
 lim
3  4 x  1 x 2  1 x3
x 

b)
2  2 x 2  1 x3
3
2

lim 4  5 x 2  x 3  4 x 4
x 

 4 5 x 2 x3 4 x 4 
 lim x 4  4  4  4  4 
x 
x
x
x 
x
 4 5 1

 lim x 4  4  2   4 
x 
x
x
x

     0  0  0  4
4
     4 
4
 
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.
MAT1581
Mathematics 1 (Engineering)
25
MODULE 6
LEARNING UNIT 3
Differentiation
The derivative
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
CONTENTS
PAGE
1.
2.
3.
26
THE TANGENT QUESTION AND THE DERIVATIVE .................................. 27
DIFFERENTIATION FROM FIRST PRINCIPLES ........................................... 28
RESPONSE TO ACTIVITY ................................................................................ 31
MAT1581
Mathematics 1 (Engineering)
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 y y f  x  h   f  x 
mAB 


change in 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
f  x  h  f  x
lim
h 0
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
d
considered on its own is called the differentiating operator, and
dx
indicates that any function written after it is to be differentiated with respect to x. We
The symbol
MAT1581
Mathematics 1 (Engineering)
27
Module 6 Learning unit 3
DIFFERENTIATION: The derivative
always differentiate with respect to the independent variable, that is
d  dependent variable 
.
d  independent variable 
Note:
dy y

dx x
y
is the average rate of change over an interval (slope of the secant) while
x
dy
is the limiting value equal to the instantaneous rate of change at a point
dx
(slope of the tangent).
The following symbols are often used to indicate derivatives instead of
f  x  h  f  x
lim
:
h 0
h
dy
i 
 we say dy by dx 
 ii  f '  x   we say f prime x 
dx
d
 f  x  
 iv  y '
 iii 
dx 
 v  Dx y
 vi  Df  x 
dy
has various interpretations:
dx
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 lim
h 0
f  x  h  f  x
h
. We will
take four steps to calculate this special limit:
28
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 3
DIFFERENTIATION: The derivative
1.
Find f  x  h
2.
Find f  x  h  f  x 
3.
4.
Divide by h
Let h  0 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:
 x2
f  x
Thus f  x  h 
Step 1:
  x  h
2
 x 2  2 xh  h 2

  
f  x  h   f  x   x 2  2 xh  h 2  x 2
Step 2:
 2 xh  h 2
f  x  h  f  x
Step 3:
h
Step 4: lim
h 0
f  x  h  f  x
h
2 xh  h 2

h
h 2x  h

h
 2x  h
 lim  2 x  h 
h 0
 2x
Thus the derivative =
f  x  h  f  x
dy
= lim
 2x
dx h0
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):
dy
 2(3)  6 .
dx x 3
Where
dy
is a shorthand notation for the derivative at x = 3.
dx x 3
MAT1581
Mathematics 1 (Engineering)
29
Module 6 Learning unit 3
DIFFERENTIATION: The derivative
We can say that the slope or gradient of the curve at x = 3 is 6.
Example 2
If f  x   2 x 2 find f '  x  from first principles.
1
Solution
Step 1: f  x  h 
 2  x  h
1
2
The factor  x  h  2 may be expanded by using the binomial theorem
1
 1  1   2 2
 1

1
    x h

 2  2 
1

2
2
 2 x  2 x h
 ..... 
2




3
1
1
3

 1  1  
 2 x 2  x 2 h       x 2 h 2  ....higher order terms in h
 2  2 
1
1
3
1

 1  1  
Step 2: f  x  h   f  x   2 x 2  x 2 h       x 2 h 2  .....  2 x 2
 2  2 
x

1
3
1
1 
2 h       x 2 h 2  ....higher order terms in h
 

 2  2 
 1  1   1  3

h  x 2       x 2 h  ....higher order terms in h 


 2  2 
f  x  h  f  x

Step 3:
 
h
h
x

1
3
1
1 
2       x 2 h  ....higher order terms in h
 

 2  2 
 1  1   1  3

f  x  h  f  x
Step 4: lim
 lim  x 2       x 2 h..  ....higher order terms in h 

h 0
h 0 
h
 2  2 


All the terms containing an h become 0.
x
30

1
2
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 3
DIFFERENTIATION: The derivative
ACTIVITY 1
Determine the derivatives of the following from first principles:
a) y  3x 2 at x  2
b)
f  x    2x  2
3
1
x
8
d) y  x (Hint: Use the binomial theorem.)
Remember to check the response on page 31.
c)
3.
f  x  x 
RESPONSES TO ACTIVITY
Activity 1
a)
Put y  f  x   3x 2
Step 1:
f  x  h
 3 x  h
2

 3 x 2  2 xh  h 2

 3x 2  6 xh  3h 2

  
Step 2: f  x  h   f  x   3x 2  6 xh  3h 2  3x 2
 6 xh  3h 2
Step 3:
f  x  h  f  x
6 xh  3h 2

h
h
h  6 x  3h 

h
 6 x  3h
f  x  h  f  x
 lim  6 x  h 
h 0
h 0
h
 6x
Step 4: lim
dy
 6  2   12
dx x   2
MAT1581
Mathematics 1 (Engineering)
31
Module 6 Learning unit 3
DIFFERENTIATION: The derivative
f  x
b)
  2x  2
3
 8 x3  24 x 2  24 x  8
f  x  h
Step 1:
  2  x  h   2 
3
 8  x  h   24  x  h   24  x  h   8
3
2
 8 x3  24 x 2 h  24 xh 2  8h3  24 x 2  48 xh  24h 2  24 x  24h  8

Step 2: f  x  h   f  x   8 x3  24 x 2 h  24 xh 2  8h3  24 x 2  48 xh  24h 2  24 x  24h  8

 8 x 3  24 x 2  24 x  8

 24 x 2 h  24 xh 2  8h3  48 xh  24h 2  24h
Step 3:
f  x  h  f  x
h


h 24 x 2  24 xh  8h 2  48 x  24h  24
h
 24 x  24 xh  8h  48 x  24h  24
2
Step 4: lim
h 0

f  x  h  f  x
2
h 0
h

 lim 24 x 2  24 xh  8h 2  48 x  24h  24

 24 x 2  48 x  24
 6  2x  2
2
Thus f '  x   24 x 2  48 x  24  6  2 x  2 
2
c)
f  x  x 
Step 1:
1
x
f  x  h   x  h 
1
 x  h
1
1
x
xh
x
1
1
h

xh x
xh  x  h   x  x  h

x  x  h
Step 2: f  x  h   f  x    x  h  

32
x 2 h  xh 2  h
x  x  h
MAT1581
Mathematics 1 (Engineering)

Module 6 Learning unit 3
DIFFERENTIATION: The derivative
Step 3:
f  x  h  f  x
h


Step 4: lim


x 2 h  xh 2  h
h
x  x  h
x  x  h
h
x 2  xh  1
x  x  h
f  x  h  f  x
h 0
1
h x 2  xh  1
h
x 2  xh  1
 lim
h 0 x  x  h 
x2  1
x2
1
 1 2
x

f ' x  1 
1
x2
Put y  f  x   x8
d)
Step 1: f  x  h 
  x  h
8
 x8  8 x 7 h  28 x 6 h 2  .....
Step 2: f  x  h   f  x   x8  8 x 7 h  28 x 6 h 2  .....  x8
 8 x 7 h  28 x 6 h 2  ....higher order terms in h
Step 3:
f  x  h  f  x
h


h 8 x 7  28 x 6 h  ....higher order terms in h
h
 8 x  28 x h  ....higher order terms in h
7
Step 4: lim
6
f  x  h  f  x
h 0

h

 lim 8 x 7  28 x 6 h  ....higher order terms in h
h 0

 8x7
Thus
dy
 8x7 .
dx
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.
MAT1581
Mathematics 1 (Engineering)
33
MODULE 6
LEARNING UNIT 4
DIFFERENTIATION
Standard forms
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.
CONTENTS
PAGE
1.
2.
2.1
2.2
34
STANDARD FORMS .......................................................................................... 35
RESPONSES TO ACTIVITIES ........................................................................... 39
Activity 1 .............................................................................................................. 39
Activity 2 .............................................................................................................. 39
MAT1581
Mathematics 1 (Engineering)
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.
y or f  x 
1
constant, k
dy
or f '  x 
dx
0
2
x
1
3
xn
nx n1
n is a constant
4
ax n
anx n1
a and n are constants
5
ex
ex
6
e kx
kekx
k is a constant
7
ax
a x n a
a is a constant
8
n x
1
x
9
log a x
1
a is a constant
x n a
10
sin kx
k cos kx
k is a constant
11
cos kx
 k sin kx
k is a constant
12
tan kx
k sec2 kx
k is a constant
13
cot kx
k cos ec2 kx
k is a constant
14
sec kx
k sec kx tan kx
k is a constant
15
cos ec kx
k cos ec kx.cot kx
k is a constant
MAT1581
Mathematics 1 (Engineering)
35
Module 6 Learning unit 4
DIFFERENTIATION: Standard forms
Example 1
Differentiate
a)
b)
y=6
y = 6x
Solution
a)
6 is a constant. Use standard form 1.
dy
 0.
dx
b)
Use standard form 4 with a = 6 and n = 1.
dy
  6 1 x11  6 x 0  6
dx
Example 2
Find the derivatives of
y  12 x3
12
b) y  3
x
c) y  3 x
a)
Solution
dy
 anx n 1 .
dx
Use standard form 4: If y  axn then
a)
b)
c)
a = 12 and n = 3 thus
dy
 12  3 x 31  36 x 2 .
dx
12
is rewritten in the standard form ax n as y  12 x 3 . So a = 12 and n =
3
x
dy
36
3, thus
 12  3 x 31  36 x 4   4
dx
x
y
1
y  3x 2 .
y  3 x is rewritten in the standard form ax as
n
1
1
 1  2 1 3  2
3
x
 x 
dy
1
   
, thus
dx
2
2
2
3
1
2x 2

3
2 x
So a = 3 and n =
.
Example 3
Find f '  x  if
a)
f  x   2x
b)
f  x   n x
c)
f  x   e3x
d)
f  x   log2 x
Solution
a)
Refer to standard form 7 with a = 2. Thus f '  x   2 x n 2 .
1
b)
Refer to standard form 8. Thus f '  x  
c)
Refer to standard form 6. Thus f '  x   3e3 x .
36
x
.
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 4
DIFFERENTIATION: Standard forms
d)
Refer to standard form 9. Thus f '  x  
Example 4
dy
Determine
if:
dx
a)
y  sin x
b)
y  sec3x
x
y  cos
4
c)
1
.
xn 2
Solution
a)
b)
c)
dy
 cos x
dx
dy
Refer to standard form 14. k  3 Thus
 3sec 3 x tan 3 x
dx
dy
1
x
Refer to standard form 11. k  14 Thus
  sin
dx
4
4
Refer to standard form 10. k  1 Thus
ACTIVITY 1
Determine the derivatives to x of the following:
a)
x5
b)
sin 3 x
c)
cot 2 x
d)
3
e)
log10 x
f)
n 3 x
h)
5 x6
5 x
x
g)
e
i)
tan 3 x
j)
10 x
k)
cos ecx
l)
log 3 x
m)
3
x4
n)
6x10
x4
p)
4
Remember to check the response on page 39
o)
sec
x
2
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)
y  t   sin5t
b)
y  t   n3t
c)
f  s  5 s
Solution
MAT1581
Mathematics 1 (Engineering)
37
Module 6 Learning unit 4
DIFFERENTIATION: Standard forms
a)
y '  t   5cos 5t
b)
y 't  
c)
1 
1
1
f  s   5 s  s 5 Thus f '  s   s 5  4 
5
55 s4
5s 5
(3) 1

3t t
1
4
ACTIVITY 2
1. Determine the derivatives of the following:
y  t   cos ec 5t
a)
b)
g  t   6t 3
c)
f s  s
d)
u  w   n w
e)
V  h   log 3 h
2. Differentiate the following:
a)
y  n 2
b)
f  x   x7
3
c)
d)
r  h2
1
v
f)
t
1
w 2
u
f t   x
g)
f  x  x 2
e)
Remember to check the responses on pages 39 and 40.
38
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 4
DIFFERENTIATION: Standard forms
2.
RESPONSES TO ACTIVITIES
2.1
Activity 1
a)
5x4
b)
3 cos 3x
c)
2 co sec 2 2 x
2
g)
1 3
1
1
x  2 
3
33 x2
3x 3
1
xn10
1
3x
5e 5 x
h)
30 x 5
i)
3sec 2 3 x
j)
k)
10 x n10
 cos ecx.cot x
l)
1
x n 3
d)
e)
f)
1
m)
o)
3 4
x
4
1
4 x3  x3
4
 
2.2
Activity 2
1.
a)
y '  t   5cos ec 5t.cot 5t
b)
g '  t   18t 2
n)
60x 9
p)
1
x
x
sec tan
2
2
2
1
c)
d)
e)
1 
1
f ' s  s 2 
2
2 s
1
u '  w 
w
1
V 'h 
h n 3
MAT1581
Mathematics 1 (Engineering)
39
Module 6 Learning unit 4
DIFFERENTIATION: Standard forms
2.
a)
b)
dy
 0 , because n 2 is a constant. You can find the value with your
dx
calculator.
f '  x   7 x6
1
c)
dr 3 32 1 3 2
 h  h . In this case r is the dependent variable and h the
dh 2
2
independent variable.

1
2 Thus
1
3
dv
1  1
1 
 t 2  t 2.
dt
2
2
d)
vt
e)
w
f)
f  t   x . Be careful here - the independent variable is given as t; thus x must
1
dw
2
 u 2 . Thus
 2u 3  3 .
2
du
u
u
be regarded as a constant in this case.
f 't   0
g)
f '  x   2 x 2 1
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.
40
MAT1581
Mathematics 1 (Engineering)
MODULE 6
DIFFERENTIATION
Rules of differentiation I
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
CONTENTS
PAGE
1.
1.1
1.2
1.3
1.4
2.
2.1
2.2
2.3
2.4
2.5
41
DERIVATIVES OF COMBINED EXPRESSIONS ............................................ 42
Constant times a function...................................................................................... 42
Sums and differences ............................................................................................ 42
Product rule ........................................................................................................... 45
Quotient rule ......................................................................................................... 46
RESPONSES TO ACTIVITIES ........................................................................... 47
Activity 1 .............................................................................................................. 47
Activity 2 .............................................................................................................. 47
Activity 3 .............................................................................................................. 48
Activity 4 .............................................................................................................. 49
Activity 5 .............................................................................................................. 49
MAT1581
Mathematics 1 (Engineering)
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 F  x   k f  x  then F '  x   k f '  x  where k is a constant.
That is, the derivative of k times a function is simply k times the derivative of the
function.
Example 1
a)
b)
c)
 
If g  x   7 x then g '  x   7  5 x   35 x
If f  x   9 x 3 then f '  x   9 3x 2  27 x 2
5
6
6
5
5
3
2 2
dy 2  5 2 1  5 2
  x  x
If y  x then
 3
3
dx 3  2

ACTIVITY 1
Find the derivatives of
a)
f  x   3x 2
b)
y  12 x 4
c)
V  2 y3
Remember to check the response on page 47.
1.2
Sums and differences
If F  x   f  x   g  x  then F '  x   f '  x   g '  x 
If F  x   f  x   g  x  then 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.
42
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation I
Example 2
Find the derivatives of
a)
F  x   x 2  sin x
b)
f  x   7 x5  4 x3  2 x  8
c)
y  e 2 x  n x  x
Solution
Differentiate the given functions term-by-term using the correct standard forms:
a)
F '  x   2 x  cos x
b)
f  x   35 x 4  12 x 2  2
c)
y  2e
1
2x
1 1 
1
1
  x 2  2e 2 x  
x 2
x 2 x
ACTIVITY 2
Determine
dy
a)
if y  3x 2  5 x  6
dx
b)
f '  x  if f  x   2 x 
4
x

d  x2 3
c)
  2  x
dx  2 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
3t 2  4t 5
a)
S
t3
b)
y   3x  2 
2
Solution
3t 2  4t 5 3t 2 4t 5
a) S 
 3  3
t3
t
t
2
1
 3t  4t
MAT1581
Mathematics 1 (Engineering)
43
Module 6 Learning unit 5
DIFFERENTIATION: Rules of differentiation I



 
ds
  3 1 t 11   4  2  t 21
dt
 3t 2  8t
3 8t
 2 
1
t
3  8t 3

t2
8t 3  3

t2
If possible, we simplify answers to the same format as the original question.
b) y   3x  2 
2
 9 x 2  12 x  4

dy
  9  2  x 21  12 1 x11  0
dx
(Remember x 0  1)
 18 x  12
 6  3x  2 
ACTIVITY 3
1. Differentiate with respect to the independent variable:
d
d  z2 z7 
a)
4  3x  2 x 2
b)
  
7
dx
dz  2


2 3

x x2
c)
y
e)
y  2x 4  4x 4
3
1
d)
f  t   2t 4  3t 3
f)
y
3
2
2
x

2
x
dy
in the following cases:
dx
a  bx  cx 2
a)
y
x
a
b)
y  ax 
ax
Remember to check the response on page 48.
2.
44
Find
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation I
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 F  x   f  x  g  x  then 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
dy
Determine
if y   3 x  1 2 x  1
dx
Solution
Let f  x   3 x  1 and g  x   2 x  1
f '  x   3 and

g ' x  2
dy
 f  x g ' x  f ' x g  x
dx
  3 x  1 2    3 2 x  1
 6x  2  6x  3
 12 x  5
Example 5
If y  x 2 sin x find
dy
.
dx
Solution
Let f  x   x 2 and g  x   sin x
dy
 x 2  cos x    2 x  sin x 
dx
 x 2 cos x  2 x sin x
ACTIVITY 4
Determine the derivatives of
a)
y   3 x  5 x 2  2 x

b)

f  t   t cos t
c)
y  xe 2 x
Remember to check the response on page 49.
MAT1581
Mathematics 1 (Engineering)
45
Module 6 Learning unit 5
DIFFERENTIATION: Rules of differentiation I
1.4
Quotient rule
The quotient rule is given by the following formula:
f  x
If F  x  
g  x
then F '  x  
g  x f ' x  f  x g ' x
 g  x   2
Consider the following examples:
Example 6
sin x
dy
If y  x  
find
.
x
dx
Solution
Let f  x   sin x and g  x   x
Then f '  x   cos x and g '  x   1
Thus
g  x f ' x  f  x g ' x
dy

2
dx
 g  x  
 x  cos x    sin x  (1)

 x 2

x cos x  sin x
x2
Example 7
Find y ' if y 
x 1
x 1
Solution
Let
f  x 
1
x  1  x 2 +1 and g
 x
1
1
1 2
1 
x and g '  x   x 2
2
2
g  x f ' x  f  x g ' x
Then f '  x  
Thus y ' 
1
 x 2 -1
 g  x  
2
 x  1 x    x  1 x 

 x  1
 x  x  x  x 

 x  1
1
1
2
2
 12
1
1
1
2
1 1
2
2
1
2
2
2
 12
1 1
2
2
1
2
1
46
 12
1
2
2
1
2
 12
2
2
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation I
1
1
1  12
 12  12 x  2
22 x

2
1
 x  1
2
 x 2
1

 x  1
1
2
2
1

x
 x  1
2
ACTIVITY 5
Determine the derivatives of
2  x2
a)
y
2  3x
et
f t   2
b)
t 2
tan x
y
c)
2x
x
f ( x) 
d)
n x
Remember to check the response on page 49.
2.
RESPONSES TO ACTIVITIES
2.1
Activity 1
a)
f '  x   32 x 
c)
 
V  2 3 y 
2.2
Activity 2
b)
a)
y  12 4 x 3
2
y  3x 2  5 x  6
dy
   3 2  x 21   51 x11  0
dx
 6x  5
b) f  x   2 x 
4
x
 2x 2  4x 2
1
1
MAT1581
Mathematics 1 (Engineering)
47
Module 6 Learning unit 5
DIFFERENTIATION: Rules of differentiation I
1 1
 1 1
 f '  x    2    x 2 1   4     x  2 1
2
 2
 x 2  2 x 2
3
1
1
2
 3
1
x2 x2
x2
 3
x2
x2

x3
x2

x x
2
 d  x2

3
d x
2
c)


x

    3x  x 
2
dx  2 x
 dx  2


2 x 21

 3  2  x 21  1 x11
2
 x  6 x 3  1
6
 x  3 1
x
2.3
Activity 3
1.
a)
3  4x
b)
z  z6
c)
d)
e)
f)
2.
a)
b)
48
dy
2 6
 2  3
dx
x
x
3  14
1
f '  t   t  2t  3
2
dy 3  14
5
 x  x 4
dx 2
dy
1
1


dx 4 x x x
dy
a
c 2
dx
x
dy
a
a


dx 2 ax 2 x ax
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation I
2.4
a)
Activity 4

dy
  3 x  5  .  2 x  2   3. x 2  2 x
dx
 6 x 2  4 x  10  3 x 2  6 x

 9 x 2  2 x  10
1
b)
c)
1 
f '  t   t   sin t   t 2 .cos t
2
cos t
  t .sin t 
2 t
dy
 x (2e2 x )  (1)( e2 x )
dx
 2 xe2 x  e2 x
 e 2 x  2 x  1
2.5
Activity 5
a)
y
2  x2
2  3x
Let f ( x )  2  x 2 and g ( x )  2  3x
f '  x   2 x
g '  x  3

dy  2  3x  2 x   2  x
 
dx
 2  3x  2


2
  3
4 x  6 x 2  6  3x 2
 2  3x  2
3x 2  4 x  6

 2  3x  2
3x 2  4 x  6
 2  3x  2
MAT1581
Mathematics 1 (Engineering)
49
Module 6 Learning unit 5
DIFFERENTIATION: Rules of differentiation I
b)
et
 2
t 2
f t 
 t  2  e    e   2t 
f ' t  
t  2
 e  t  2  2t 

t  2
t
2
2
2
t
2
2
2
c)
y 
tan x
2x
t


2
dy  2 x  sec x   tan x  2 

dx
 2 x 2


2 x sec 2 x  tan x

2
4x
x sec x  tan x

2 x2
2
d)
f ( x) 
x
n x
f ' x 

 n x 1  x 
 n x  2
1

 x
n x  1
 n x  2
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.
50
MAT1581
Mathematics 1 (Engineering)
MODULE 6
LEARNING UNIT 6
DIFFERENTIATION
Rules of
differentiation II
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
CONTENTS
PAGE
1.
1.1
1.2
2.
2.1
2.2
2.3
2.4
2.5
51
DERIVATIVES OF COMPOSITE FUNCTIONS .............................................. 52
Function-of-a-function or chain rule..................................................................... 52
General standard forms ......................................................................................... 61
RESPONSES TO ACTIVITIES ........................................................................... 65
Activity 1 .............................................................................................................. 65
Activity 2 .............................................................................................................. 65
Activity 3 .............................................................................................................. 66
Activity 4 .............................................................................................................. 66
Activity 5 .............................................................................................................. 68
MAT1581
Mathematics 1 (Engineering)
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 y  x   x3 and x  t   sin t , we can write y   sin t  . There are two
3
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 F  x   f  g ( x)  then 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 y  y ( x ) where x  x (t ) then
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
4
dy
Find
if y  x 3  1 .
dx
Solution
Let g ( x)  x3  1


Then from the chain rule
dy
dx
 f '  g ( x)  . g '( x)

  3x 
 4 x3  1
3
2
This answer can be simplified.


=12x 2 x 3  1
52
3
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
Example 2
If y (t )   sin t  find
3
dy
.
dt
Solution
Make the substitution x  t   sin t
Thus y  x3 .
Now differentiate each of the functions:
dy
dx
 3 x 2 and
 cos t
dx
dt
Then using the chain rule we can write:
dy
dy dx
 
dt
dx dt
 
 3x 2  cos t
Remove the substitution by substituting x  sin t .
 3  sin t  cos t
2
 3sin 2 t cos t
Example 3
Differentiate y  1  x2 .
Solution
We note that y  1  x2 is a function, namely the square root of another function,
1  x2 .
We can make a substitution:

1

Let u  1  x 2 then y  u  u 2
To use the chain rule we require
du
dy
.
and
dx
du
1
dy 1  2
du
and
=  2x
 u
du 2
dx
Thus
dy
dx

dy du

du dx

1 2
u  2 x 
2
1

1
2
  xu
Substitute u  1  x 2

MAT1581
Mathematics 1 (Engineering)

53
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II

  x 1  x2




1
2
x
1
2 2
1 x


x
1  x2
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 2
.
y  4 1 x

Solution
3
dy 4 3
d
2 2 1


1  x2
 . 1  x 

dx 1 2
dx





1
 6 1  x 2 2  2 x 


1
 12 x 1  x 2 2
Example 5
2
dy
1

Find
if y      .

d

Solution
2
2
1

S          1 




2 1 d
dy
 2    1 
   1
d
d

 2    1  1   1 2 
1
1 
1 

 2    1  2 
   

1 1 1 

 2     3 
   

1 

 2   3 
 

54
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
ACTIVITY 1
Differentiate with respect to the independent variable:
8
1
s
f  y   y3  y 2
a)
b)
1 t2


c)
s  x 1
d)
f  t   2  3t 2
e)
y  3 4  9x
f)
y


3
1
3
x 3
2
3
b 

y   a  2  a and b are constants
x 

Remember to check the response on page 65.
g)
In the following example we apply the product rule as well as the function-of-afunction rule:
Example 6
ds
Find
if s  3t 2  2
dt

 5t  4.
Solution
s  3t 2  2

 5t  4
  3t  2   5t  4 
2
1
2
We will use a substitution for the product rule but not for the function-of-a-function
part.
Function
of a function


1
2
Let u  3t  2 and v   5t  4  2
du
 6t
dt
and
1 1 d
dv 1
 5t  4 2
 5t  4 
dt 2
dt

1
1
 5t  4  2  5 
2

5
1
 5t  4  2
2
Writing the product rule in
d
notation we have
dt
ds
dv
du
 u.  v.
dt
dt
dt

 52  5t  4 
 3t 2  2 .
MAT1581
Mathematics 1 (Engineering)
 12
  5t  4  2 .  6t 
1
55
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
There are various methods to simplify. We will illustrate two methods:
Method 1
1
ds
5
1
 3t 2  2 .  5t  4  2   5t  4  2 .  6t 
dt
2


Take out the highest common factor (HCF) in both terms:



1
1
1
1
 5t  4  2 5 3t 2  2  12t  5t  4  2  2 
2

HCF



1
1
 5t  4  2 15t 2  10  60t 2  48t
2

75t 2  48t  10
2. 5t  4
Method 2
1
ds
5
1
 3t 2  2  5t  4  2   5t  4  2  6t 
dt
2


Write the first term as a fraction. Add terms using rules for handling fractions.


1
5
   5t  4  2  6t 
 3t 2  2 
1
 2  5t  4  2 







5 3t 2  2  12t  5t  4  2


1 1
2
2  5t  4  2
1
15t 2  10  60t 2  48t
2  5t  4  2
1
75t 2  48t  10
2. 5t  4
ACTIVITY 2
Differentiate with respect to x and simplify:
5
a) y  x 1  2 x 
b)
r  x 2 ax  b
c) s   2 x  1 1  x
d)
r   2 3  4
e)   x x 2  a 2
f)
y  x4  2 x3  7


3
4
Remember to check the response on page 65.
56
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
In the following examples we will use the quotient rule and the chain rule.
Example 7
If y 
x3
 x  1
dy
and simplify.
dx
find
4
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
4
y  x 3  x  1 .]
Function
of a function

 
dy  x  1 3 x  x . 4  x  1 1

2
dx
 x  14 


4

2
3
3
H.C.F.


3
2
x  x  1 3  x  1  4 x 
 x  1
8
x 2  x  1  3 x  3  4 x 
3


 x  1
8
x2 3  x 
 x  15
Example 8
Determine f '  x  if f  x  
Solution
f  x 



 and simplify the answer.
3 2x2  1
1  x2

3 2 x2  1
1  x2


3 2 x2  1
1  x 
2
1
2
Function
of a function


1  x  .3  4 x   3  2 x  1 . 1  x   2 x 
 f ' x 
 1 x
 

2
1
2
2
2
MAT1581
Mathematics 1 (Engineering)
1
2
1
1
2  2
2
2
57
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
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
1
denominators in the top and bottom. LCM in this example is 1 x2  2 .

12 x 1 x 2

1
2
1
 f ' x 


3 x 2 x 2 1

1 x 
2
2


1
2
1


1 x 
2
2


 1 x 
3 x 2 x 2 1

2 2
1 x 
1 x  1 x 
1

1
1
2

2
1  x 
12 x 1 x 2

1

2
1
1
2
1
2
1
2
1
  3x  2 x  1
1  x 
12 x 1  x
2 1
2
1
2 1 2
3x  4  4 x 2  2 x 2  1
1  x 
3x  3  2 x 

1  x 
2
3
2
2
2 3
Example 9
1  x2
Find the derivative of y 
using
x
quotient rule
a)
b)
product rule
Solution
a)

x. 12 1  x 2
dy

dx


 . 2 x   1. 1  x
 12
2
x2
x2

1 x 2
2
1  x2
x
 x 1 x
2
 12x
2
2
x

58
1
x2 . 1  x2
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
1
1  x2
2
 x 1 1  x 2
x
 
y
b)

Function
of a function


1
 12
dy
1
2
  x 1
1  x2
 2 x   1  x 2  x 2
dx
2


  

1
1  x 


1  x2
2

1

1
 

2
x2
2
 
x 1  x 
1 1
2
2
x2  1  x2
2


2
1
2
x2 1  x2

x2 1  x2

1
2
2x2 1
x2 1  x2
ACTIVITY 3
1. Differentiate with respect to the independent variable and simplify:
2x2  x  1
3
y 2
r
a)
b)
2s  1
3x  x  1
x 2
c)
y
2
Differentiate by using (i) the quotient rule and
d)
x3
y
x 2  3x
x5
(ii) the product rule
 x 
y

 1 x 
a)
5
4
 x3  1 
b)
y 3 
 2x 1 
Remember to check the response on page 66.
Example 10
By making a substitution u  x 2  3x , use the chain rule to find the derivative of
y  n x 2  3 x .


Solution
MAT1581
Mathematics 1 (Engineering)
59
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
Let u  x 2  3x then y  n u
In this example the chain rule becomes
Now
du
dy 1
 2 x  3 and

dx
du u
Thus
dy dy du


dx du dx

1
  2 x  3
u

2x  3
x 2  3x
dy dy du
.


dx du dx
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
y  n f  x  . We have
If y  n f  x  then
dy f '  x 
.

dx
f  x
Example 11
By making a substitution u  x 2  3x , use the chain rule to find the derivative of
y  ex 3x .
2
Solution
Let u  x 2  3x then y  eu
In this example the chain rule becomes
Now
du
dy
 2 x  3 and
 eu
dx
du
Thus
dy dy du


dx du dx
dy dy du
.


dx du dx
 eu   2 x  3 
  2 x  3 e x  3 x
The result of the previous example can be generalised to any function of the form
2
y  e   . We have
f x
If y  e   then
f x
60
dy
 f '( x) e f ( x ) .
dx
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
Example 12
By making a substitution u  x 2  3x , use the chain rule to find the derivative of
y  sin x 2  3 x .


Solution
Let u  x 2  3x then y  sin u
du
dy
Now
 2 x  3 and
 cos u
dx
du
dy dy du
Thus


  2 x  3 cos u
dx du dx
The result of the previous example can be generalised to any function of the form
y  sin 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
dy
dx
y
n 1
1
 f  x  
2
e f (x
f ' x e  
3
a  
f '  x  a   n a
4
n f  x 
f ' x
5
loga f  x
1 f ' x
.
n a f  x 
6
sin f  x 
f '  x  cos f  x 
7
cos f  x 
 f '  x  sin f  x 
8
tan f  x 
f '  x  sec2 f  x 
9
cot f  x 
 f '  x  cos ec2 f  x 
10
sec f  x 
f '  x  sec f  x  tan f  x 
11
cosec f  x 
 f '  x  cosec f  x  .cot f  x 
n
f x
MAT1581
Mathematics 1 (Engineering)
n  f  x  
. f ' x
f x
f x
f  x
61
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
ACTIVITY 4
Differentiate the following:


a)
y  n x 2  1
b)
y  n  2 x  5 
c)
y  n  n x 
d)
y  e3 x 2
e)
y  e 2  e 3 x  n 3
f)
y  log  2 x  1
g)
y  4 2 x 5 x
h)
y  esin 3 x
i)
y  105 x
j)
y  log 2 3 x 4
2
2
3
2
x
k)
y  x2  1 . e 2
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
 1  x2 
dy
if y  n 
.
2 
dx
1 x 
Solution
 1  x2 
  n 1  x 2  n 1  x 2
y  n 
2 
1 x 
Find





a


log b  log a  log b 
2x
2 x
dy


2
dx 1  x 1  x 2

   
1  x 1  x 
2 x 1  x 2  2 x 1  x 2
2
2

2 x  2 x 3  2 x  2 x 3
1  x4

4 x
1  x4
Example 14
Find the first derivative if y 
tan x  1
.
sec x
Solution
Simplify first: y 
62
tan x  1
sec x
sin x
1
x
cos

1
cos x
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
1
 sin x  cos x 


cos x

 cos x
 sin x  cos x  cos x


cos x
1


 sin x  cos x
The differentiation becomes very easy:
dy
 cos x    sin x 
dx
 cos x  sin x
What would happen if we did not simplify first?
tan x  1
y
sec x


dy sec x sec x   tan x  1 sec x tan x 
 
dx
sec 2 x

2
sec3 x  tan 2 x.sec x  sec x.tan x
sec 2 x
sin 2 x
sin x
1
. 1  cos1 x . cos
 cos
2
x
cos3 x
x cos x

1
cos 2 x

1  sin 2 x  sin x.cos x
cos x
 Multiply top and bottom by cos x 
3
sin 2 x  cos 2 x  1




 cos 2 x  1  sin 2 x 
cos 2 x  sin x cos x

cos x

cos x  cos x  sin x 
cos x
 cos x  sin 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:
a)
y   x  2
c)
y
MAT1581
Mathematics 1 (Engineering)
4
1
4

2
2x
x

b)
y  2 x2  4 x  5
d)
y

6
a2  x2
a2  x2
63
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
a2  x2
a2  x2
w
e)
y
g)
y
i)
y  4 x 3  x  3 2
1  4 w2
1
f)
y  x 3  2 x2
h)
y  1 x
j)
y  x 2 1  x2
3


1
2
2. Differentiate with respect to the independent variable and simplify the
answer:
a)
R  log3  4 x  1
b)
p  cos  n x 
c)
y  cot e2 x
d)
s  esin x
2
3. Find the value of f '  x  for the given value of x  in radians  .

a)
f  x   n tan x ; x 
b)
f  x   5e 2 sin
c)
f  x   10e 10 sin 3x; x  1
x
x
2
4
; x2
x
4. Find the first derivative of:
a)
y  sin 4 x
b)
y  n cos x
c)
f  x   sin  2 x  2 
e)
y  cos3 x 2
f)
y
g)
y  tan x 2
h)
y  tan 2 x
i)
y  tan 2 3x 4
j)
f  x   x 2 sin x
 
d)  x.sin x
1  cos x
1  cos x
5. Differentiate with respect to the independent variable, and simplify
where possible:
a)
y  3sin x 2  4
b)
y  sin 3  3x  3


c)
y  sin x
d)
  sin x
e)
  cos3 x 2
f)
y
1
tan x.sin 2 x
2
Remember to check the response on pages 68 and 69.
64
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
2.
RESPONSES TO ACTIVITIES
2.1
Activity 1
a)
S
1
 1  t 2 
 12
1 t2
 1 2 1 d
ds
1
   1  t 2 
1 t2
2
dt
dt
3
 2
1
  1 t2
 2t 
2
t

3
2
1 t2







t
1  t 
2 3
Only the answers to the rest of the questions are supplied.

 3y  2 y 
b)
f '  y   8 y3  y2
d)
f '  t   18t 2  3t 2
f)
dy

dx

7
2

2
2 x
33
 x  3
2
4
2.2
Activity 2
a)
dy
4
 1  2 x  1  12 x 
dx
c)
ds
3  6x

dx 2 1  x
e)
d 2 x 2  a 2

dx
x2  a2
MAT1581
Mathematics 1 (Engineering)
c)
ds
1

dx 2 x  1
e)
3
dy

dx 3  4  9 x 2
g)
dy
b 
6b 
  3 a  2 
dx
x 
x 
b)
dr 5ax 2  4bx

dx
2 ax  b
d)
dr 2 3  5

d
3  4
f)
3 4 x3  6 x2
dy

dx 4. 4 x 4  2 x 3  7


2
2


65
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
2.3
Activity 3
In questions a) to c) it is not necessary to use the chain rule.
5x  x  2
dy
dr
6
1.
a)
b)


2
dx
ds  2 s  12
3x 2  x  1




c)
dy  x  3 x

dx
x3
2.
a)
dy
5 x 4

dx 1  x 6
2.4
Activity 4
a)
This function has the form y  n f  x  with f  x   x 2  1
Therefore
b)
3 x 2  45 x  20
dy

dx 2  x  5 2 2  3 x
b)
2
3
dy 36 x x  1

5
dx
2 x3  1




3
dy
2x
 2
dx x  1
Given y  n  2 x  5 
Note that
d)
3
y  3 n  2 x  5 
dy
2
  3
dx
 2 x  5

c)
6
 2 x  5
y  n  n x  is of the form y  n f  x  where f  x   n x
Therefore
f ' x
dy

dx
f  x
1
 
x
 
n x
1
1
 
x n x

66
1
x n x
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
d)
2
dy
 6 xe3 x  2
dx
e)
dy
 0  3e3 x  0  3e3 x (Since e2 and n 3 are constants.)
dx
f)
Use the table of standard forms, number 5, with a = 10 and f  x   2x  1.
dy
1
2

.
dx log10 2 x  1

g)
2
2x  1
(Since log 10 = 1)
Use the table of standard forms, number 3, with a = 4 and f  x   2 x2  5 x .
2
dy
  4 x  5  4 2 x  5 x .  n 4 
dx

  n 4  4 x  5  42 x 5 x
2

It is important to use brackets in your answers to prevent confusion.
h)
Use the table of standard forms, number 6.
dy
  3cos 3 x  esin 3 x
dx

i)



2
dy
 (10 x) 105 x  n10 
dx

Using a calculator this can be simplified to 23,03 x 105 x
2
.
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)
dy
1 12 x3
.

dx n 2 3x 4

k)
y
4
x n 2

 .e
x 2  1.e 2 x  x 2  1
1
MAT1581
Mathematics 1 (Engineering)
1
2
x
2
67
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II

 .e    e    x  1  2 x 
e  x  1
xe


2
 x  1
e  x  1  2 xe

2  x  1
e  x  1  2x

2  x  1
e  x  2 x  1

1
 y '  x2  1
x
x
2
1
2
2
x
1
2
2
2
x
1
2
2
x
1
2
x
2
2
2
2
2
1
2
x
 12
2
2
1
2
x
1
2
2
2
2
2
2 x2  1
2.5
Activity 5
1.
a)
y '  4  x  2
c)
y'  
e)
y' 
g)
y' 
i)
y '  2 x 2  x  3
3
1
2

3
x
x3
2a 2 x
a  x  a  x
2
2
4
4
1
1  4w 
2 3
 12
 7 x  18

b)
y '  24  x  1 2 x 2  4 x  5
d)
y' 
f)
y' 
h)
y' 
j)
y '  12 x 2 1  x 2
dp  sin  n x 

dx
x
2.
a)
dR 4 log 3 e

dx
4x 1
b)
c)
dy
 2e 2 x cosec 2 e 2 x
dx
d)
a)
 
f '   1
4
b)
c)
f ' 1  27

5
4a 2 x
a  x 
2 2
2
3  4x2
3  2 x2
1
4 xx x
1

  5 x  3
 12
2
2
ds
 2 x cos x 2 esin x
dx
3.
68
f '  2   21.35
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 6
DIFFERENTIATION: Rules of differentiation II
4.
a)
dy
 4 cos x.sin 3 x
dx
b)
dy
  tan x
dx
c)
f '  x   2 cos  2 x  2 
d)
du
 x.cos x  sin x
dx
e)
y  cos3 x 2  cos x 2
f)
dy
2 sin x

dx 1  cos x  2
h)
dy
 2 tan x sec 2 x
dx
j)
f '  x   x 2 cos x  2 x sin x
b)
dy
 9 sin 2  2 x  3 cos  2 x  3
dx
  



3
2
dy
 6 x sin x 2 cos x 2
dx
g)
dy
 2 x sec 2 x 2
dx
i)
dy
 24 x 3 tan 3x 4 sec2 3x 4
dx
 
 
5.


a)
dy
 6 x cos x 2  4
dx
c)
dy cos x

dx
2 x
d)
cos x
d

dx 2 sin x
e)
d
 6 x cos2 x 2 .sin x 2
dx
f)
dy
 2sin x cos x  sin 2 x
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.
MAT1581
Mathematics 1 (Engineering)
69
MODULE 6
LEARNING UNIT 7
DIFFERENTIATION
Higher order
derivatives
OUTCOMES
At the end of this learning unit, you should be able to repeatedly differentiate a
function to find higher order derivatives.
CONTENTS
PAGE
1.
2.
70
HIGHER ORDER DERIVATIVES ..................................................................... 71
RESPONSES TO ACTIVITY .............................................................................. 77
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives
1.
HIGHER ORDER DERIVATIVES
dy
or f '  x  is again a function of x on its own, which in turn may be
dx
differentiated.
The function
dy
 15 x 2 and 15x 2 is a function that can be
dx
differentiated with respect to x.
For example if y  5 x3 , then
Thus if y  5 x3 , then
d  dy  d
15 x 2  30 x.
 
dx  dx  dx


The expression [30x] is called the second differential coefficient (second derivative)
d2 y
d  dy 
of the original function, and is denoted by the symbol 2 which means
 .
dx  dx 
dx
d2y
dy
Thus
measures the rate at which
is changing with respect to x,
2
dx
dx
dy
just as
measures the rate at which y is changing with respect to x.
dx
d2 y
can also be differentiated with respect to x,
dx 2
and the result is the third differential coefficient (third derivative) of y with respect to x.
Consequently
It is represented by the symbol
d3 y
.
dx 3
Thus in the above example:
 5 x3
Function:
y
dy
 15 x 2
First derivative:
dx
d2y
Second derivative: 2  30 x
dx
d3y
 30
Third derivative:
dx 3
d4y
Fourth derivative: 4  0
dx
As seen above, the repeated differential soon becomes 0 in some cases. In other cases
it may continue indefinitely.
MAT1581
Mathematics 1 (Engineering)
71
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives
Example 1
Find the repeated differential coefficients of y  x3  x 2  x  1 .
dy
 3x 2  2 x  1
dx
d2 y
 6x  2
dx 2
d3 y
6
dx3
d4 y
 0, as well as all the higher differential coefficients
dx 4
Example 2
Find the repeated differential coefficients of y 
1 1
.

x x2
y  x 1  x 2
dy
 1.x 2  2.x 3
dx
d2 y
 2 x 3  6 x 4
2
dx
d3 y
 6 x 4  24 x 5
3
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
d2 y
dx 2
d3 y
dx3
d4 y
dx 4

f '  x
y'
Dy
 '  x
f ''  x 
y ''
D2 y
 ''  x 
f '''  x 
y '''
D3 y
 '''  x 
f 'v  x 
y 'v
D4 y
 'v  x 




Second derivative
Third derivative
Fourth derivative

72
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives
The successive differential coefficients are read as follows:
dy
dy by dx
dx
d2 y
d two y by dx two
dx 2
d3 y
d three y by dx three, etc.
dx3
2
Note: There is a big difference between
d2y
 dy 
and   .
2
dx
 dx 
3
There is also a big difference between
d3y
 dy 
and   .
3
dx
 dx 
The next example will illustrate the difference:
Example 3
If y  1  x  x 2 , determine
3
d3y
 dy 
and   .
3
dx
 dx 
Solution
y  1  x  x2
dy
 1  2 x
dx
d2y
2
dx 2
d3 y
0
dx3
3
3
 dy 
and     1  2 x 
 dx 
d 3 y  dy 
Clearly
 
dx3  dx 
3
Example 4
Determine f '''  x  if f  x  
2
, simplify and give your answer with positive indices.
x 1
2
Solution
You can use the quotient rule or rewrite f(x) to use the product rule.
1
2
f  x  2
 2 x2  1
x 1

MAT1581
Mathematics 1 (Engineering)

73
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives

  2x
2
f '  x   2  1 x 2  1

4 x
 
2
 4 x x 2  1
 x  1
2

2
  2 x    x  1  4  Using the product rule 
 16 x  x  1  4  x  1
3
f ''  x   4 x  2  x 2  1
2

3
2
16 x 2
2
2
2
3
2

2
2
 x  1  x  1
16 x  4  x  1

 x  1
2
2
4

3
2
2
12 x 2  4
 x  1
4  3x  1

 x  1
3
2
2
3
2


 x  1
 x  1  4  6 x   4  3x  1  3  x  1  2 x 
f '''  x  
 x 1 
4 3x2  1
2
3
2
3
2
2
2
3 2
2
 Using the quotient rule 
 x  1  24 x    24 x   3x  1 x  1

 x  1
x  1  x  1  24 x    24 x   3 x  1 




 x  1
 x  1  24 x    24 x   3 x  1 


 x  1
3
2
2
5
2
2
2
2
2
2
6
2
2
4
2

24 x3  24 x  72 x 3  24 x
 x  1
2
74
2
2
4
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives

48 x 3  48 x
 x  1
48 x  x  1

 x  1
4
2
2
2
4
Example 5
If u  1  2 x  3x 2 , determine an expression for
2
d 2u
d 2  du 
2
u    2.
2
dx
dx
 dx 
 
Solution
Find the values of the derivatives first, then substitute into the given expression and
simplify.
du
 2  6x
dx
2
2
 du 
2
    2  6 x   4  24 x  36 x
 dx 
d 2u
6
dx 2
 




2
d 2
d
 2 1  2 x  3x 2  2  6 x 
u 
1  2 x  3x 2
dx
dx

  4  12 x  1  2 x  3 x 2

 4  8 x  12 x 2  12 x  24 x 2  36 x 3
 4  20 x  36 x 2  36 x 3
 
d 2u
d 2
Now:
u
2
2
dx
dx

 du 
 
 dx 
2
 
2

 6  2 4  20 x  36 x 2  36 x3  4  24 x  36 x 2  2
 6  8  40 x  72 x 2  72 x3  4  24 x  36 x 2  2
 4  16 x  36 x 2  72 x 3
MAT1581
Mathematics 1 (Engineering)
75
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives
ACTIVITY 1
1. Determine y ' in a) to d):
1
a)
y
x2  9
x
b)
y
x 1
x2  2
x
c)
y
d)
 2 x  3x 
y
2
2.
a)
b)
1
2
x 1
d2y
Determine
if:
dx 2
a  bx
y
a  bx
x
y
2x 1
3.
Determine f '''  x  if:
a)
f  x   3x 4  8 x3  12 x 2  5
b)
f  x 
c)
1
4 x
f  x   a  bx
4.
Find the values of y' and y'' for the given values of x :
a)
y  ax 
b)
y  x x 2  9;
5.
1 ds 1 2 d 2 s
If s  7t show that t  t
 6s
2 dt 10 dt 2
6.
If y  3sin(2 x  3) show that
7.
a2
; xa
ax
x4
6
d2y
 4y  0
dx 2
If y  sin x  2cos x show that y ''' y '' y ' y  0
Remember to check the response on page 77.
76
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives
2.
RESPONSES TO ACTIVITY
Activity 1
1.
a)
x
y'  
 x  9
2
b)
y' 
c)
y' 
x2
2  2 x  1
2.
3.
4.
2
x2  2
1  4x
 x  1  2 x  3x 2 
2
a)
d2y
4ab 2

dx 2  a  bx 3
b)
d 2 y   x  2

5
dx 2  2 x  1 2
a)
f '''  x   24  3 x  2 
b)
f '''  x  
c)
f '''  x  
a)
b)
5.
y' 
3
2
x
d)
3
1
2
6
4  x
4
3b3
8  a  bx  2
5
y'  0 

x  a
 y ''  1 
2a 

41 

y'  5 
x  4

 y ''  236 
125 

s  7t 6
ds
 42t 5
dt
d 2s
 210t 4
dt
Now left-hand side
MAT1581
Mathematics 1 (Engineering)

1 ds 1 2 d 2 s
t  t
2 dt 10 dt 2
77
Module 6 Learning unit 7
DIFFERENTIATION: Higher order derivatives



1
1
t 42t 5  t 2 210t 4
2
10
6
6
 21t  21t


 42t 6
 
 6 7t 6
 6s
 Right-hand side
6.
y  3sin  2x  3
dy
 6 cos  2 x  3
dx
d2 y
 12sin  2 x  3
dx 2
Left-hand side
=
d2y
 4y
dx 2
 12sin  2 x  3  4  3sin  2 x  3 
0
 Right-hand side
7.
y  sin x  2 cos x
y '  cos x  2sin x
y ''   sin x  2 cos x
y '''   cos x  2sin x
y  y ' y '' y '''  0
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.
78
MAT1581
Mathematics 1 (Engineering)
MODULE 6
LEARNING UNIT 8
DIFFERENTIATION
Applications I
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
CONTENTS
PAGE
1.
2.
2.1
2.2
3.
3.1
3.2
4.
4.1
4.2
4.3
79
L’HOSPITAL’S RULE ........................................................................................ 80
THE GRADIENT OF A CURVE ......................................................................... 81
The equation of a tangent to a curve ..................................................................... 81
The equation of a normal to a curve ..................................................................... 83
RATE OF CHANGE ............................................................................................ 84
Velocity ................................................................................................................. 84
Acceleration .......................................................................................................... 84
RESPONSES TO ACTIVITIES ........................................................................... 86
Activity 1 .............................................................................................................. 86
Activity 2 .............................................................................................................. 87
Activity 3 .............................................................................................................. 87
MAT1581
Mathematics 1 (Engineering)
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:
If lim
f  x
is of the form
f  x
f ' x
0
then lim
 lim
x a g  x 
x a g '  x 
0
 x
f ' x
if lim
is meaningful.
xa g '  x 
xa g
In some cases L’Hospital’s rule must be applied more than once until the denominator
is not 0.
Example 1
x  tan 2 x
x 0 x  tan 2 x
x  tan 2 x
0
It is clear that the lim
yields the indeterminate form .
x 0 x  tan 2 x
0
Evaluate lim
Therefore we apply L'Hospital's rule:
x  tan 2 x
1  2sec 2 2 x
 lim
x 0 x  tan 2 x
x 0 1  2sec 2 2 x
1 2

1 2
 3
lim
Example 2
e 2 t  et  1
2e 2 t  et

lim
t 0
t 0
2t
t2
0
which is still in the form 0  apply rule again
Evaluate lim
4e 2 t  e t
t 0
2
3

2
 lim
80
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
ACTIVITY 1
Determine
x2  1
x 4  7 x 2  12
a) lim
b) lim
x 1 x  1
x2
x2
2
sin 
c) lim
 0 1  cos 2 
Remember to check the response on page 86.
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 y  x 2 at the point x  3 .
dy
dy
Slope =  2 x and
 6.
dx
dx x 3
Example 4
At which point on the curve y 
dy
2
 2 x 2  2
dx
x
Put the gradient = -2
2
2  2
x
2
2 x  2
2
is the gradient to the curve equal to –2?
x
The gradient =
x2  1
x  1
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 
MAT1581
Mathematics 1 (Engineering)
81
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
Example 5
Find the equation of the tangent to the curve of y =3x2 + 4x + 5
at the point (3; 10).
Solution
Method 1
In this case a  3 and b  10.
Let
f  x   3 x 2  4 x  5
 f '  x   6 x  4
 f '  3  6  3  4
 14
Equation of tangent is
y   10   14  x  3
Substitute values in 1 
y  10  14 x  42
 y  14 x  32
OR
14 x  y  32  0
Method 2
We can also use the gradient-intercept form of the straight line: y  
mx 
Gradient
c .

y-intercept
Let the equation of the tangent be y  mx  c .
Now y  3 x 2  4 x  5
dy
 6 x  4
dx
If x  3, the gradient of the tangent: m  6  3  4
 14
The tangent is passing through the point (3; 10), therefore these values must satisfy
the equation.
 10  14  3  c
10  42  c
 c  32
 Equation of tangent: y  14 x  32
82
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
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).
1
because the conditions for two
f ' a
lines to be perpendicular to each other is that the product of their gradients equals –1.
Hence the gradient of the normal to y is equal to
The equation of the normal at P  a, b is thus
Equation of the normal: y  b 
1
 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
Let
f  x   3 x 2  4 x  5
 f '  x   6 x  4
 f '  3  6  3  4
 14
Equation of normal is
1
y   10   
 x  3
14
x 3
 y  10  
14 14
x 3
 y    10
14 14
x 143
 
14 14
OR
x  14 y  143  0
ACTIVITY 2
1. Determine the equation of (a) the tangent and (b) the normal to the
curve x 2  y 2  25 at the point (4 , 3) on the curve.
2x
2. Find the equation of the tangent to the curve y  2
at the origin.
x 1
Remember to check the response on page 87.
MAT1581
Mathematics 1 (Engineering)
83
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
3.
RATE OF CHANGE
dy
is the instantaneous rate of change of y with respect to x for a
dx
definite value of x.
We have said
dy
dy
 8.
 2 x . When x = 4 then
dx
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.
For example, for the function y  x 2 we have
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  .
s
is the average velocity when P moves a distance  s in the time  t . In the
t
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  t  0 .
Now
Velocity  v 
ds
dt
Thus velocity is the rate of change of displacement with respect to time.
3.2
Acceleration
Acceleration = Rate of change of velocity with respect to time
dv
dt
d  ds 
  
dt  dt 
=
d 2s
 2
dt
84
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
Example 7
A body moves according to the equation s  25t  t 2 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
s  25t  t 2
ds
 25  2t
Velocity 
dt
 When t  3, the velocity  25  2  3  19 m/s
Example 8
The distance s metres moved by a body in t seconds is given by
s  2t 3  13t 2  24t  10 .
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
a)
s  2t 3  13t 2  24t  10
ds
Velocity 
 6t 2  26t  24
dt
When t  4, the velocity  6  4   26  4   24
2
 96  104  24 m/s
 16 m/s
b)
When the body comes to rest
ds
0
dt
 6t 2  26t  24  0
 3t 2  13t  12  0
  3t  4  t  3  0
 t  3 or 43
 The values of t when the body comes to rest are 3 seconds or 1 13 seconds.
MAT1581
Mathematics 1 (Engineering)
85
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
ds
 6t 2  26t  24
dt
d 2s
Acceleration  2  12t  26
dt
Acceleration is given as 10 m/s 2
Velocity 
c)
12t  26=10

12t =36

t =3
 The value of t when the acceleration is 10 m/s 2 is t  3 seconds.
ACTIVITY 3
1.
A missile is fired into the air. The height h in metres of the missile
after t seconds is given by h  19.2t  4.8t 2 .
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?
Remember to check the response on page 87.
4.
RESPONSES TO ACTIVITIES
4.1
Activity 1
x2  1
2x
 lim 
x 1 x  1
x 1
1
 2  1
a)
lim
b)
 2
x  7 x  12
4 x3  14 x
 lim
x 2
x 2
x2
1
4
2
lim
 4  2   14  2 
3
 32  28
4
86
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
c)
sin 2 
2sin  .cos
 lim
2
 0 1  cos 
 0 2 cos  .   sin  
lim
2sin  .cos
 0 2 cos  .sin 
 lim 1
 lim
 0
1
4.2
Activity 2
1.
a)
x 2  y 2  25

y  25  x  25  x
2

dy 1
 25  x 2
dx 2

2
1
2
  2 x  

1
2
x
25  x
dy
4
Gradient of tangent at  4;3 :

dx
3
4
Equation of tangent: y  3    x  4 
3
3 y  9  4 x  16
4 x  3 y  25  0
b)
Gradient of normal  
Equation of normal:
2

x
y
1
3

4
 3  4
3
 x  4
4
4 y  12  3 x  12
3x  4 y  0
y 3
2.
The equation of the tangent at the origin is y  2 x
4.3
Activity 3
1.
a)
b)
2.
19.2 m/s
14.4 m
ds
 3t 2  8t  5
Velocity  v 
dt
d 2 s dv
 6t  8
Acceleration: 2 
dt
dt
2
Stationary after 1 and 1 seconds
3
2
Acceleration  2 m/s and 2 m/s 2
MAT1581
Mathematics 1 (Engineering)
87
Module 6 Learning unit 8
DIFFERENTIATION: Applications I
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.
88
MAT1581
Mathematics 1 (Engineering)
MODULE 6
LEARNING UNIT 9
DIFFERENTIATION
Applications II:
maxima and minima
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
CONTENTS
PAGE
1.
1.1
1.2
1.3
2.
3.
4.
4.1
4.2
89
MAXIMA AND MINIMA ................................................................................... 89
Definitions............................................................................................................. 89
Maximum value .................................................................................................... 90
Minimum value ..................................................................................................... 90
DERIVED CURVES ............................................................................................ 91
PRACTICAL APPLICATIONS ........................................................................... 99
RESPONSES TO ACTIVITIES ......................................................................... 103
Activity 1 ............................................................................................................ 103
Activity 2 ............................................................................................................ 107
MAT1581
Mathematics 1 (Engineering)
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
dy
derivative at that point. This means that
 0 at point M and m.
dx
dy
 0 , we can find the value of x where y is a maximum or
dx
minimum. To distinguish between a maximum and minimum, the change of the
curve in the vicinity of these points must be studied.
If we solve the equation
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.
90
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and 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
dy
shape of the slope curve near a value of x that makes
 0.
dx
Now the slope of this new curve at this point is clearly negative, that is
negative when y reaches a maximum value. Thus
1.3
d  dy 
  is
dx  dx 
d2y
 0 when y is a maximum.
dx 2
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
dy
shape of the curve of the slope near a value of x that makes
 0.
dx
Now the slope of this new curve at the point is clearly positive, that is
d  dy 
  is
dx  dx 
d2 y
positive when a minimum is reached. Thus
 0 when y is a minimum.
dx 2
MAT1581
Mathematics 1 (Engineering)
91
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Figure 3
The following procedure is used to find maximum or minimum values:
dy
 Put
 0 to find the possible values of x.
dx
d2 y
d2 y
 Find
.
Substitute
the
values
of
x
into
.
dx 2
dx 2
d2 y
If
 0 the function reaches a maximum at that point.
dx 2
d2 y
If
 0 the function reaches a minimum at that point.
dx 2
2.
DERIVED CURVES
Consider the graph of y  f  x  in figure 4.
Figure 4
At point A the function reaches a maximum.
92
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
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
dy
 0 at the turning
dx
points.
MAT1581
Mathematics 1 (Engineering)
93
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
The graph of the second derivative y  f ''  x  shows clearly that when
d2y
 0;
dx 2
d2y
y  f  x  is a minimum, f ''  x   2  0;
dx
d2y
y  f  x  has an inflection point, f ''  x   2  0.
dx
y  f  x  is a maximum, f ''  x  
This information allows the value of the turning points to be determined by
dy
differentiating the function and solving the equation
 0 for x.
dx
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
d2 y
determined by considering 2 .
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 y  x3  6 x 2  9 x  2 and sketch the curve.
Solution
(1)
dy
0
dx
y  x3  6 x 2  9 x  2
dy
 3 x 2  12 x  9
dx
dy
0
For critical values
dx
 3 x 2  12 x  9  0
Turning points occur at
x2  4 x  3  0
 x  1 x  3  0
 x  1 and x  3
We need to find the nature of these points.
(2)
94
To determine the nature of these points substitute x  1 and x  3 into
d2 y
.
dx 2
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
dy
d2 y
 3 x 2  12 x  9 Thus
 6 x  12 .
dx
dx 2
d2 y
 6 x  12
dx 2
d2 y
At x  1:
 6 1  12
dx 2
 6  0
 negative
Thus a maximum value at x = 1.
To find the maximum value substitute x = 1 in the original equation.
 ymax  1  6 1  9 1  2
3
2
 1 6  9  2
2
Thus the coordinates of the maximum turning point are (1;2).
d2y
 6  3  12
dx 2
6
At x  3 :
 positive
Thus a minimum value at x = 3.
To find the minimum value substitute x = 3 in the original equation.
ymin   3  6  3  9  3  2
3
2
 27  54  27  1
 2
Thus the coordinates of the minimum turning point are (3; -2).
We are ready to sketch the curve:
Figure 6
MAT1581
Mathematics 1 (Engineering)
95
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Example 2
Find the turning points of the graph of f  x  
3x 2  1
and establish which is the
x  23
minimum and maximum. Sketch the graph.
Solution
f  x 
3x 2  1
x  32
 x  32  6 x   3x 2  1 1
f ' x 
2
 x  32 


6 x 2  4 x  3x 2  1
 x  32 
2
3x 2  4 x  1
 x  32 
2
For critical values put f '  x   0

3x 2  4 x  1
 x  32 
2
0
2


 Provided that x  3  0 
 3x 2  4 x  1  0
 3x  1 x  1  0
 x  1 and x 
1
3
The nature of the turning points:
3x 2  4 x  1
f ' x 
2
 x  23 
 x  32   6 x  4    3x 2  4 x  1 2  x  32  1
2
f ''  x  
 x  2 2 
3


2
 1   2   0  0 positive
f '' 1  3


1 4
3
2
At x  1:
 Minimum value at x  1
3 1 2 3
ymin  f 1  1  
1 1
3
6
 Minimum turning point 1; 6 
96
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
  1   2    13  43  1 2   13 
f ''  x   3
2
  13 
2
1
At x  :
3

 92  0
1
9
2 9
 
9 1
 2
 0  negative 
 Maximum value at x 
1
3
3
 1 1
1
ymax  f    1 91
3
3

 32
 13

2 3

3 1
2
1 
 Maximum turning point  ; 2 
3 
Note: f  x  is undefined when x  32 . Points like these need further investigation as
shown in figure 7.
Figure 7
MAT1581
Mathematics 1 (Engineering)
97
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Example 3
Find the coordinates of the maximum and/or minimum turning points of the curve
y  x.e x .
Solution
y  xe x
dy
 x.e x 1  e x 1
dx
 e x  x  1
dy
0
dx
 e x  x  1  0
For critical values
e x  0; x  1
Nature of the point
d2 y
 e x 1   x  1 e x 1
dx 2
 e x 1  x  1
 ex  x  2
If x  1:
d2y
 e 1  1  2 
2
dx
1
 1
e
 0  positive 
 Minimum value when x  1
ymin   1 e 1
1
e
 0,368

 Coordinates of minimum turning point  1;  0,368 
Note: The expression e x is always positive for any value of x. Thus e x  0.
d2 y
Also, it is not necessary to find the actual value of
.
dx 2
98
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Example 4
Find the maximum and/or minimum values of y 
x
and sketch the graph.
x 1
2
Solution
y
x
x 1
2


x2  1 1  x  2 x
dy
 
2
dx
x2  1



1  x2
 x  1
2
2
Turning points
1  x2

0
2
x2  1


1  x 2  0
 x2  1
 x  1


2
The point to note is that the denominator x 2  1 can be ignored in solving
dy
 0 as
dx
it may never equal 0.
There is no value we can give to x except x  1 or  1 that will make
dy
 0.
dx
Nature of the turning points
d y  x  1   2 x   1  x   2  x  1  2 x

dx
 x  1
2
2
2
2
2
2
2
4
d 2 y 2   2   0 negative value


 negative value
positive value
dx 2
24
 y a maximum.
2
If x  1,
d 2 y 2   2   0 positive value


 positive value
positive value
dx 2
24
 y a minimum.
2
If x  1,
1
1

11 2
1
1

Minimum value of y 
11
2
Maximum value of y 
Sketch:
MAT1581
Mathematics 1 (Engineering)
99
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Figure 8
ACTIVITY 1
1. Find the maximum and/or minimum turning points of the curve
2
y   x  5   x  1 .
2. Determine the coordinates of the maximum and/or minimum points of
the curve y  e x  e x .
3. Find the coordinates of the maximum and minimum points of the curve
defined by y  12n x  x 2  10 x.
4. Find y  sin x  cos x for 0  x   .
5. Sketch the curve of y   x  1 x  2 x  3
6. Sketch f  x    x3  2 x 2  11x  12 .
7. Sketch the curve y  x3  3x  14 .
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.
100
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Height of container
 x cm
Width of container
  28  2 x  cm
Length of container
  60  2 x  cm
V  x  28  2 x  60  2 x 
Volume of container
 1 680 x  176 x 2  4 x3
dV
 1 680  352 x  12 x 2
dx
dV
0
dx
12 x 2  352 x  1 680  0
For critical values
3x 2  88 x  420  0
 3x  70  x  6   0
 x  6 and
70
3
d 2V
 352  24 x
dx 2
If x  6 :
d 2V
 352  24  6 
dx 2
 352  144
 208
= Negative
 Maximum when x  6
 Dimensions of largest open container:
Height = 6 cm
Width = 28  2  6   16 cm
Length = 60  2  6   48 cm
Example 6
The cost of manufacture Rc, per km of an electric cable is given by c 
120
 600 x,
x
where x is its cross-section in cm 2 . Find:
a)
the cross-section for which the cost is least
b)
the least cost per km
Solution
MAT1581
Mathematics 1 (Engineering)
101
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
120
 600 x ............................. 1
x
dc
120
   2  600
dx
x
dc
For critical values
0
dx
120
 2  600  0
x
 600 x 2  120
120
 x2 
600
 x   0.2
 0.447
c
The negative root has no meaning in this connection, and is disregarded.
d 2 c 240
 3
dx 2
x
d 2c
240
When x  0.447; 2 
dx
 0.447 3
 0  positive 
Minimum when x = 0.447
a)
b)
Cost is a minimum for a cross-section of 0.447 cm2
Substituting for x = 0.447 in (1), we get the minimum cost.
120
 600  0.447 
0.447
 R 536.66
Thus c  R
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
 A  x 2  4 xh ................................... 1
[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.
102
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
 x2 h  8
h 
8
x2
substitute in 1
4x 8
.
1 x2
32
 x2 
x
A  x2 
 Now we have two variables only
32
dA
 2x  2
dx
x
dA
For critical values
0
dx
32
 2x  2  0
x
 2 x3  32
 x3  16
 x  3 16
 2.52
Nature of values:
d2A
64
 2 3
2
dx
x
2
d A
64
 2
If x  2.52 :
2
dx
 2.52 3
 positive 
0
 Minimum when x  2,52
h
8
 2.52 
2
 1.26
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.
MAT1581
Mathematics 1 (Engineering)
103
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
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.
RESPONSES TO ACTIVITIES
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.
y  sin x  cos x
4.
dy

 cos x  sin x
dx
Turning points:
 cos x  sin x  0
 sin x  cos x
sin x

1
cos x
that is
tan x  1
x 

4
Nature of points
d2y
Now:
  sin x  cos x
dx 2

d2y
1
1
If x  then


 negative
2
4
dx
2
2
 y is a maximum
104
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Maximum value of y  sin


4
 cos

4
2
2
 2
5.
Remove brackets to find
y  x3  4 x 2  x  6
Turning points
dy
 3x 2  8 x  1
dx
dy
Let
 0 for extreme values
dx
 3x 2  8x  1  0
8  64  12
6
8  7, 21

6
 2.54 or 0.13
x 
Nature of the turning points
d2y
 6x  8
dx 2
d2y
If x  2.54 : then
 15.24  8
dx 2
 7.24
0
 Graph has a minimum at x  2.54
 Value of y   2.54   4  2.54    2.54   6
3
2
 16.39  25.81  2.54  6
 0.88
If x  0.13 : then
d2y
 6  0.13  8
dx 2
 0.78  8
 7.22
0
 Graph has a minimum at x  0.13
 Value of y   0.13  4  0.13   0.13  6
3
2
 0.0022  0.0676  6.13
 6.06
MAT1581
Mathematics 1 (Engineering)
105
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
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 y   x  1 x  2 x  3
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: y  x3  4 x 2  x  6
f)
f  x    x3  2 x 2  11x  12
 f '  x   3 x 2  4 x  11
f ''  x   6 x  4
Turning points
Solve f '  x   0
3 x 2  4 x  11  0
4  16  132
6
4  12.17

6
16.17
8.17
or

6
6
or  1.36
 2.70
x
Nature of the turning points
106
MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
For x  2.70
f ''  x   6  2.70   4  0
 x  2.70 is a point of maximum value.
Value of y    2.70   2  2.70   11  2.70   12
3
2
 19.68  14.58  29.70  12
 12.6
For x  1.36 f ''  x   6  1.36   4  0
 x  1.36 is a point of minimum value.
Value of y   1.36   2 1.36   111.36   12
3
2
 2.52  3.70  14.96  12
 20.74
y-intercept: y = 12
Sketch:
Figure 10
g)
y  x3  3x  14
y-intercept = 14
MAT1581
Mathematics 1 (Engineering)
107
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
Turning points
dy
 3x 2  3
dx
dy
has no real roots since x 2  1 has no real solution.
dx
Therefore there are no turning points.
Points of inflection
d2y
 6x
dx 2
 x  0 is a point of inflection.
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.
Length
108
1 200  3x
2
3
 600  x
2

MAT1581
Mathematics 1 (Engineering)
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
3 

A  x  600  x 
2 

Area of field:
 600 x 
3x 2
2
dA
 600  3x
dx
dA
0
dx
 600  3x  0
 3x  600
For critical values
 x  200
d2A
 3
dx 2
0
 negative 
 Maximum when x  200
3 200
 600  
2 1
 300
Length
Dimensions of largest field
Length
Width
2.
=
=
300 m
200 m
s  120t  16t 2
Maximum height will occur where
ds
 0.
dt
ds
0
dt
120  32t  0
32t  120
t  3, 75 seconds
s  120t  16t 2
 120  3, 75   16  3, 75 
2
 225 metres
Check with second derivative test that it is a maximum:
d 2s
 32  0  Maximum
dt 2
MAT1581
Mathematics 1 (Engineering)
109
Module 6 Learning unit 9
DIFFERENTIATION: Applications II: maxima and minima
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.
110
MAT1581
Mathematics 1 (Engineering)
POST-TEST: DIFFERENTIATION
1.
a)
b)
c)
d)
2.
Given that f ( x)  4 x 2  3x  2 , find the values of
f (2)  f (1)
f (3  a)
f (3  a)  f (3)
f (3  a )  f (3)
a
2 x
Given g ( x)  x 2  5 x and f ( x) 
, determine
x2
a)
g  0
b)
f  0
c)
a if g  a   f  2  0
d)
y if y  f  y   0
3.
a)
Calculate the following limits:
lim y  3
b)
lim
c)
d)
e)
f)
g)
4.
y 9
 t  3 t  7 
t 2  t  1 t  4 
x2  2 x  8
x 4 x 2  5 x  4
x4  a4
lim 2
x a x  a 2
x 2  5x  3
lim 4
x 0 x  7 x  5
8x2  2 x  3
lim 3
x  2 x  3 x  1
3x  5
lim
x  2 x  3
lim
Find the following one-sided limits:
1
1
lim
and lim
x2 x  2
x2 x  2
5.
Determine the derivative of f  x   x3  3x 2  2 x  1 from first principles.
6.
Find
dy
2
from first principles if y  2 .
dx
x
7.
Find
da
if a  b 2 x 2  bx  10 x .
db
111
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test
8.
9.
Determine the gradient of the curve x = 2y2 at the point where x = 8.
dy
, simplify and write the answers with positive indices and in
dx
surd (root) form.
Determine


3
a)
y  4 1  x2
b)
y  3 1  x3
c)
y  1  x  x2
3
10.
64
dy

If y   6t 2  2  determine
.
dt
t 

11.
At what point (coordinates) on the curve of y 
2
3 x  1 is the gradient of
3
the curve equal to 1?
12.
Determine the gradient of the tangent to the curve f  t  
t2  3
at the point
t 5
on the curve where t  4. Give your answer in simplified surd form.
13.
14.
15.
Differentiate with respect to x, and simplify your answers to a single term,
with positive indices and in surd form where possible:
a)
   x  1 x 2  2 x  2
c)
1 x
y
x 1
e)
r
b)
d)
y  x p  ax  b 
y
q
x2  1
x2  4
x 1
x 1
3x 2  1
at the
Determine the gradient of the tangent to the curve f  x  
x +5
 1 
point 1;
 . Give your answer in simplified surd form.
 3 2
Find the x value(s) of the point on the curve y 
x2
where the gradient of
x4
the curve equals 0. (Hint: x  4  0 ).
16.
If f  x  
1  x  , determine f '''  x  . Simplify your answer to a single term,
2 3
and in surd form.
112
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
17.
Differentiate and simplify the answer to a single fraction if
3


4
x

2


x


y  n e 
  x  2  


18.
Differentiate with respect to the independent variable and simplify:
19.
20.
a)
y  1  x 2 .sin 3x
b)
  sin 3  t 2  2t 
c)
y  sin x. tan x  sec x  1
d)
  tan 
1 x 

1 x 
Determine the following:
2h  2
a)
lim
h 0
h
x  sin x
x 0
x2
b)
lim
c)
lim
x3  2 x2  4 x  3
x 1
4 x 2  5x  1
The displacement s metres of a particle from a fixed point at a time t seconds
t
10
is given by s  10  5te .

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
t3 2
 t litres of water in the tank. The person filling the tank is instructed to
3
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 y  x4  8 x2  16 .
23.
Determine the coordinates of the maximum and/or minimum turning point(s)
1
of f  x   x  and sketch the curve.
x
24.


If f  x   x 2  3 e x determine the minimum value correct to two decimal
places.
MAT1581
Mathematics 1 (Engineering)
113
Module 6
DIFFERENTIATION: Post-test
25.
Find the maximum value of y  sin 2 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 x  2,4sin  5t  0,1 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
y  e t sin 2t , 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  x  1 x  2  .
30.
4t  3t 2
where t is the angle of
3  4t
pitch of the screw. Show that maximum efficiency is 25%.
31.
Determine the maximum and minimum values of the function
2
The efficiency of a screw is given by E 
give a rough sketch of the curve y 
32.
If y  3e3 x  6e3 x show that
33.
Sketch the following curves:
a)
y  2 x 3  3 x 2  12 x
b)
114
x3
 3x  9 and
3
x3
 3x  9 .
3
d2y
 9y .
dx 2
y   x  1  x  2  x  3
2
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
DIFFERENTIATION: POST-TEST SOLUTIONS
1.
f ( x)  4 x 2  3x  2
a)
f (2)  4  2   3(2)  2
2
 16  6  2
 12
f (1)  4(1) 2  3(1)  2
 43 2
3
f (2)  f (1) 
12
3
4
b)
f (3  a )  4(3  a ) 2  3(3  a )  2
 4(9  6a  a 2 )  9  3a  2
 36  24a  4a 2  9  3a  2
 4a 2  21a  29
f (3)  4(3) 2  3(3)  2
c)
 4(9)  9  2
 29
f (3  a )  f (3)  4a 2  21a
d)
f (3  a )  f (3) 4a 2  21a

a
a

a (4a  21)
a
 4a  21
2.
g ( x)  x 2  5 x
a)
g (0)  (0) 2  5(0)  0
b)
f (0) 
f ( x) 
2x
x2
20 2
 1
02 2
MAT1581
Mathematics 1 (Engineering)
115
Module 6
DIFFERENTIATION: Post-test
c)
g ( a )  f (2)
0
22
( a 2  5a )  

22
0
( a 2  5a )   04   0
a 2  5a
0
a  a  5
0
a  0 or a  5
d)
y  f  y
0
2 y
y2
0
y
y  y  2  2  y 
0
y2  2 y  2  y
0
y2  3y  2
0
y
3  (3)2  4(1)( 2)
2
 0,56
3.
a)
or
 3,56
lim y  3
y 9
 12
2 3
b)
 t  3 t  7 
t 2  t  1 t  4 
 2  3 2  7 

 2  1 2  4 
5  9

1 6
lim

116
45
6
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
c)
x2  2 x  8
lim
x 4 x 2  5 x  4
 x  4  x  2 
 lim
x 4  x  4  x  1
4  2
4  1
6

3
2
x4  a4
lim 2
xa x  a 2
x2  a2
 lim
xa
1
2
 2a

d)
e)
f)
x2  5x  3
3

4
x 0 x  7 x  5
5
lim
8x2  2 x  3
lim
x  2 x 3  3 x  1
8
 x22  x33
x
 lim
x  2  32  13
x
x
0
g)
4.
5.
3  5x 3

x  2  3
2
x
lim
1
 
x 2 x  2
1
lim
 
x 2 x  2
lim
We must evaluate lim
h 0
Step 1:
f  x  h  f  x
h
.
f  x
 x3  3x 2  2 x  1
f  x  h
  x  h  3 x  h  2  x  h  1
3
2
 x3  3 x 2 h  3 xh 2  h3  3 x 2  6 xh  3h 2  2 x  2h  1
Step 2: f  x  h   f  x   3 x 2 h  3 xh 2  h3  6 xh  3h 2  2h
MAT1581
Mathematics 1 (Engineering)
117
Module 6
DIFFERENTIATION: Post-test
f  x  h  f  x
Step 3:
h


h 3 x 2  3 xh  6 x  3h  2

h
 3 x  3 xh  6 x  3h  2
2
Step 4: lim
f  x  h  f  x
h 0
h 0
h

 lim 3 x 2  3 xh  6 x  3h  2

 3x 2  6 x  2
Thus f '  x   3 x 2  6 x  2
6.
We must evaluate lim
f  x  h  f  x
h 0
Put y 
f  x 
2
x2
f  x  h
Step 1:
h
 2 x 2
 2  x  h
2
Use the binomial theorem on  x  h 
f  x  h

.
2

 2  3 x 4 h2  .....
 2  x  2   2  x  3  h  

1.2


 
 2 x  2  4 x  3h  6h 2 x 4  .....
Step 2: f  x  h   f  x 
Step 3:
f  x  h  f  x
h
 4 x  3 h  6 x 4 h 2  .....terms containing higher orders of h
 4 x  3  6 xh  terms containing higher orders of h
f  x  h  f  x
 lim  4 x  3  6 xh  terms containing higher orders of h
h 0
h 0
h
 4 x 3
(all terms containing h become 0)
Step 4: lim
Thus
7.
dy
 4 x  3
dx
da
 2bx 2  x
db
1
 x 2
x  2y  y   
8.
2
dy
1
Note that
.
=
dx
dx
2
 
dy
118
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions

dx
dy
 4y
dy
dx



1
4y
1
x
4 
2
1
 1
4 x
2
2
2
4 x
dy
dx x 8
Gradient at x  8 is
dy
2

dx x 8 4 8
2

4.2. 2
1
8

9.
a)

  1  x 
dy 3
 1  x   2 x 
dx 4
3
y  4 1  x2
2


b)
2
1
3
4
4
1
6 x
4
1  x2
4


6 x
4. 4 1  x 2
y  3 1  x3

 1  x3


1
3
 
2
dy 1
3
 1  x3
3 x 2
dx 3

  x 2 1  x3

2

3
 x2

3
1  x 
MAT1581
Mathematics 1 (Engineering)
3 2
119
Module 6
DIFFERENTIATION: Post-test
y  1  x  x2
c)

 1  x  x2

1

2

1
dy 1
2
 1  x  x 2  2 x  1
dx 2


10. y  6t 2  6t 2
2 x  1
2. 1  x  x 2
 
3
4
 6t 2  6t 2


3
4
 
 14
dy 3 2
 6t  6t 2
. 12t  12t 3
dt 4

 6t  6t 
9 t  t 



1
2
2

9 t t 3
1
1
4
3
1
  2 1  4
6  t  t 2  

 
1

9t  3 
t 
 
 2 1
4 6 t 


t2 

11.
1
2
2
3 x  1   3 x  1 2
3
3
dy
1
1
  3 x  1 2 .3
dx
3
y
  3 x  1

Given
dy
dx
 12
1

 3x  1
1
2
1
3x  1
 1 thus
1
3x  1
1
3x  1  1
3x  1  1
2
3
Substitute the value of x into the original equation to find the value of y.
x
120
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
y 
2 2
3   1
3 3

2
2 1
3

2
3
12. Gradient =
23
81 13
.
13.
a)
 12
d
1
 x 2  2 x  2   x  1   x 2  2 x  2
 2x  2
dx
2

 x  12
 x  2x  2 
2
x2  2x  2
2x2  4x  3

x2  2x  2
dy
q
q 1
 px p 1  ax  b   x p  q  ax  b   a 
dx
  ax  b 
c)
q 1
 px p 1  ax  b   qax p 


dy  x  1 1  1  x 1

dx
 x  12

d)
x2  2x  2
x2  2 x  2  x2  2 x  1

b)

2
 x  12

x2  4
dy

dx
  2 x    x  1    x  4   2 x 
1
2
2
1
2
2
 12
x2  4
 x  4  2 x    x  1  x 

 x  4 x  4
2
2
2

2
1
2
x3  7 x
 x  4  x  4 
2
MAT1581
Mathematics 1 (Engineering)
2
121
Module 6
DIFFERENTIATION: Post-test
dr
e)

dx
 x  1     x  1   x  1     x  1
 12
1
2
1
2
 12
x 1
 
1 x 1
 2 x 1
 2
1
1
2  1 x 1
2 x 1
 x  1
1
1 

 1  1 x 1 2  1 x 1 2 
2 x 1
x  1  2 x 1



 
 
1
1 


1

1
x
x
1
2
2



1
1


x
x
2  x  1 



   


  x  1   x  1 
1

1 
2  x  1 
   x  1 x  1  2 




1
1
 x  1   x  1 x  1  2
1
 x  1 x  1
 x  5  12   3x 2  1  6 x    3x 2  1
f ' x 
2
 x  5
 12
14.

1
2
1
2
 x  5 6 x   3x 2  1
 x  5
2
3x 2  1
15 x  1

 x  5 2 3x 2  1
15 1  1
Gradient 
1  52 3 12  1


15.
16
36 2
2 2
9
y '  x 2  1 x  4 
2
 2x  x  4
1
y'  0
122
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
 x2  x  4
2
 2x  x  4
1
0
 x2  2x  x  4  0
 x2  2 x2  8x  0
x2  8x  0
x  x  8  0
x  0 or 8
16.
3 1
2
3
f '  x     1  x2
 2 x 
2


 
1
f ''  x   3 1  x   3 x   1  x   2 x 
2
1
 3 x 1  x 2
1
2
2
1
2  2
2
 

3 1  x   3 x

1  x 
1
 3 1  x 2
 3x 2 1  x 2
2
2
2

 12
2
1
2
6 x2  3

1  x 
1  x  12 x    6 x  3   1  x   2 x 
f '''  x  
1
2
1
2
2
2
2
1
2
2
 12
1  x2
1  x  12 x   x  6 x  3

1  x 
2
2
2



17. y  x n e 
3
2
12 x  12 x 3  6 x 3  3 x

1  x2

3
2
9 x  6 x3

1  x2

3
2
9 x  6 x3
1  x  1  x
2
2
3
n  x  2   n  x  2  
4
MAT1581
Mathematics 1 (Engineering)
123
Module 6
DIFFERENTIATION: Post-test
y '  n e 
 1


3 1
1 


4  x  2 x  2 
3
3

4  x  2 4  x  2


4 x2  4  3 x  2  3 x  2
4  x  2  x  2 
x2  1
 x  4
2
18.
y  1  x 2 .sin 3 x
a)

 1  x2

 .sin 3x
1
 Product rule 
2
1
1
dy
2
2
1
 1  x 2 .cos 3x  3  sin 3x   1  x 2  2 x 
dx
2






 .cos  3x   x sin 3x
1
1  x 
3 1  x 
cos 3x  x sin 3x

1  x 
3 1  x  cos 3 x  x sin 3 x

3 1  x2
1
2
2
2
1
2
1 1
2
2
2
1
2
2
1  x2
b)
  sin 3 t 2  2t




 sin t 2  2t 



3
 Function of a function rule
31 d
d
d
 3 sin t 2  2t  . sin t 2  2t  . t 2  2t 




dt
dt
dt






2


 3 sin t 2  2t  .cos t 2  2t .  2t  2 






 6  t  1 sin  t  2t  .cos  t  2t 
2
 3  2 sin t 2  2t  .cos t 2  2t  t  1


2
124
2
2
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
c)
dy
 cos x tan x  sin x sec 2 x  sec x tan x
dx
1
1
tan x 
tan x
cos x
cos x
 sin x 
 sin x
d)
d
 1  x  1  x  1  1  x 
 sec2 
.
dx
 1 x 
1  x 2

19.
a)
2
 1 x 
sec2 

 1 x 
1  x 
2
lim
h0
2h  2
h
Substitution gives
 lim
1
2
 2  h   0
h 0
b)
1
2
1
 12 (2)

0
, use 'Hospital's rule:
0
 12
1
2 2
x  sin x
x 0
x2
lim
Substition gives
0
use 'Hospital's rule:
0
1  cos x
x 0
2x
= lim
Substition gives
= lim
x 0

0
use 'Hospital's rule again:
0
0  sin x
2
0
2
0
c)
x3  2 x 2  4 x  3
x 1
4 x2  5x  1
lim
MAT1581
Mathematics 1 (Engineering)
125
Module 6
DIFFERENTIATION: Post-test
Substition gives
0
use 'Hospital's rule:
0
3x 2  4 x  4
= lim
x 1
8x  5
20.
a)

34 4
85

3
3
1
ds
d
t

5 te  10
dt
dt


Use product rule
t
 1 t
 5e  10   5t     e  10
 10 
  1 
t
 5e  10 1   t  
  10  
b)
d 2s
t 
1 
t
t
 1
 5e  10 .  1    5e  10   
2
10  10 
dt
 10 
t 
t
t
 1
 1
 5e  10   
 5e  10   

 10 100 
 10 
t
1
t
 1
 5e  10   
 
 10 100 10 
1
t
 t
 5e  10 
 
100 5 
c)
 1 
1   10  t   0
   
1
 t  0
 10 
t  10
21.
V 
t3 2
t
3
dV
 t 2  2t
dt
2
t  2t  15
t 2  2t  15  0
 t  5 t  3  0
t 5
126
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
y  x 4  8 x 2  16
22.
y '  4 x3  16 x
4 x3  16 x  0


4 x x2  4  0
x  0; x  2; x  2
To find maximum and minimum do second derivative test.
y ''  12 x 2  16
x  0 : y ''  16  0  maximum
x  2 : y '' 
x  2 : y '' 
23.
8  0  minimum
4  0  minimum
1
x
x
1
f ' x  1  2
x
1
1 2  0
x
2
x 1
0
x2
x2  1  0
f  x 
x  1
y  2
2
x3
x  1:
f ''  x  
f ''  x   0  min
x  1: f ''  x   0  max
MAT1581
Mathematics 1 (Engineering)
127
Module 6
DIFFERENTIATION: Post-test
24.


f  x   x2  3 e x
 
 e  x  2 x  3
f '  x   e x x2  3  ex  2 x 
x
2
f ' x  0
 x  3 x  1  0
x  3 of/or x  1


f ''  x   e x x 2  2 x  3  e x  2 x  2 
x  1 : f ''  x   0  min
Min  f 1  1  3 e1  5, 44
25.
y  sin 2 x  x
y '  2 cos 2 x  1
y' 0
2 cos 2 x  1  0
1
2
2 x  1.0472
x  0.5235
cos 2 x 
y ''  4sin 2 x
y ''x 0,5235  3.464
 0  max
y  sin 2  0.5235   0.5235
 0.023
128
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
26.
Volume = x 2 h
and
Area = x 2  4 xh
1200  x 2  4 xh
1200  x 2
4x
2
x3
2 1200  x 
V x 
  300 x 
4x
4


h 
dV
3x 2
 300 
dx
4
dV
0
For maximum put
dx
3x 2
 300
4
x 2  400
x  20
Check with 2nd derivative test:
d 2V 6 x

4
dx 2
d 2V
 0 Thus a maximum value at x  20
dx 2 x  20
1200  400 
Volume of largest box = 400 

80

 4000 cm3
MAT1581
Mathematics 1 (Engineering)
129
Module 6
DIFFERENTIATION: Post-test
27
dx
 2, 4 cos  5t  0,1 5 
dt
 12 cos  5t  0,1
a)
5t  0,1 

2
  0,1
t 2
5
 0.334
Maximum displacement x  2.4sin  5  0.334   0.1
 2.4
2
Check for max:
d x
 60sin  5t  0.1
dt
 Remember the angle is in radians
d2 x
 60  0  maximum
dt t 0.334
b)
x  2.4 sin 1  0.1
 2.4 sin  0.9 
dx
 12 cos  5t  0.1
dt
 12 cos 1  0.1
 1.88 m
28.
=  7.45 m/s
y '  et sin 2t  2e t cos 2t
et   sin 2t  2 cos 2t   0
2 cos 2t  sin 2t
tan 2t  2
2t  1,107
t  0.554
max: y  e t sin 2t

sin 2  0.554 
e0.554
 0,5139
Remember second derivative test.
130
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
29.
F  x    x  1 x  2 
2
F '  x    x  2    x  1 2  x  2 
2
  x  2   x  2   2 x  2 
  x  2  3x  4 
F '  x   0  x  2 and x  
F ''  x    3x  4    x  2  3
4
3
 6 x  10
F ''  2   2  0  max
F  2    2  1 2  2 
2
0
30.
E
4t  3t 2
3  4t

dE  3  4t  4  6t   4t  3t

dt
 3  4t 2

2
  4
12  2t  24t 2  16t  12t 2
 3  4t 2


MAT1581
Mathematics 1 (Engineering)
12t 2  18t  12
 3  4t 2

6 2t 2  3t  2
 3  4t 

2
131
Module 6
DIFFERENTIATION: Post-test
E'0
2t 2  3t  2  0
 2t  1 t  2   0
t
1
of/or t  2
2
 3  4t 2  24t  18  6  2t 2  3t  2   3  4t  2  4 
E '' 
 3  4t 4
 3  4t  24t  18  48  2t 2  3t  2 

3
 3  4t 

t
E
150
 3  4t 
3
1
: E ''  0  max
2
4  12   3  12 
2
3  4  12 
5
1
4
5 4
 25%
31.
f  x 
x3
 3x  9
3
f '  x   x2  3  f '  x   0  x2  3  x   3
f ''  x   2 x
132
f ''  x  x  3  0
 min
f ''  x  x  3  0
 max
3 3
3 3 9  92 3
3
3 3
y
3 39 92 3
3
y
MAT1581
Mathematics 1 (Engineering)
Module 6
DIFFERENTIATION: Post-test solutions
32.
y '  9e3 x  18e3 x
y ''  27e3 x  54e3 x

 9 3e3 x  6e3 x

 9y
33.
y  2 x 3  3 x 2  12 x
a)
dy
 6 x 2  6 x  12
dx
2
6 x  6 x  12  0
x2  x  2  0
 x  2  x  1  0
x  2
x  1
y  20
y  7
y  0 : 2 x3  3 x 2  12  0


x 2 x 2  3 x  12  0
3  105
4
x  1.81x  3.31
x  0: x 
MAT1581
Mathematics 1 (Engineering)
133
Module 6
DIFFERENTIATION: Post-test
b
y   x  1  x  2  x  3
2
Intercepts : y  axis:
x  axis :
y  6
x  1,  2, 3
Extreme values
f '  x   2  x  1 x  2  x  3   x  1  x  3   x  1  x  2 
2
2
  x  1  2  x  2  x  3   x  1 x  3   x  1 x  2  
  x  1  4 x 2  5 x  11
Put f '  x   0
Turning points at x  1, x  1.15 or x  2.40
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.
134
MAT1581
Mathematics 1 (Engineering)
INTEGRATION
CONTENTS
LEARNING UNIT 1 REVERSE OF DIFFERENTIATION I
1.
2.
3.
3.1
3.2
3.3
THE PROCESS OF INTEGRATION ................................................................. 138
THE GENERAL SOLUTION OF INTEGRALS OF THE FORM ax n ............. 139
RESPONSES TO ACTIVITIES.......................................................................... 141
Activity 1 ............................................................................................................. 141
Activity 2 ............................................................................................................. 142
Activity 3 ............................................................................................................. 143
LEARNING UNIT 2 REVERSE OF DIFFERENTIATION II
1.
2.
3.
4.
4.1
4.2
4.3
2.
152
METHOD OF SUBSTITUTION ........................................................................ 153
RESPONSE TO ACTIVITY ............................................................................... 156
LEARNING UNIT 4 STANDARD INTEGRALS
1.
144
SIMPLIFY THE INTEGRAND BEFORE INTEGRATION ............................. 145
DETERMINE THE VALUE OF THE CONSTANT OF INTEGRATION ....... 147
APPLICATION: DISPLACEMENT, VELOCITY, ACCELERATION ............ 148
RESPONSES TO ACTIVITIES.......................................................................... 149
Activity 1 ............................................................................................................. 149
Activity 2 ............................................................................................................. 150
Activity 3 ............................................................................................................. 150
LEARNING UNIT 3 METHOD OF SUBSTITUTION
1.
2.
PAGE
137
161
TABLE OF STANDARD INTEGRALS ............................................................ 162
 f (x)
dx .......................................................... 163
INTEGRALS OF THE FORM 
 f(x)
3.
INTEGRALS OF THE FORM  f '  x  ef  x  dx .................................................. 165
4.
INTEGRALS OF THE FORM  f (x).af(x) dx .................................................. 166
5.
5.1
5.2
5.3
5.4
RESPONSES TO ACTIVITIES.......................................................................... 168
Activity 1 ............................................................................................................. 168
Activity 2 ............................................................................................................. 168
Activity 3 ............................................................................................................. 169
Activity 4 ............................................................................................................. 169
MAT1581
Mathematics I (Engineering)
135
M
O
D
U
L
E
72
Module 7
INTEGRATION: Contents
LEARNING UNIT 5 PARTIAL FRACTIONS
1.
2.
INTEGRATION USING PARTIAL FRACTIONS ............................................ 173
RESPONSE TO ACTIVITY ............................................................................... 175
LEARNING UNIT 6 TRIGONOMETRIC FUNCTIONS
1.
2.
2.1
2.2
187
DEFINITION ...................................................................................................... 188
EVALUATION OF A DEFINITE INTEGRAL ................................................. 188
RESPONSE TO ACTIVITY ............................................................................... 190
EXERCISE 1 ....................................................................................................... 191
LEARNING UNIT 8 AREAS
1.
2.
3.
4.
4.1
4.2
5.
178
INTEGRATION OF TRIGONOMETRIC FUNCTIONS .................................. 179
RESPONSES TO ACTIVITIES.......................................................................... 182
Activity 1 ............................................................................................................. 182
Activity 2 ............................................................................................................. 184
LEARNING UNIT 7 THE DEFINITE INTEGRAL
1.
2.
3.
4.
172
193
INTRODUCTION ............................................................................................... 194
SUMMATION AND THE DEFINITE INTEGRAL .......................................... 195
THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE ................. 196
RESPONSES TO ACTIVITIES.......................................................................... 205
Activity 1 ............................................................................................................. 205
Activity 2 ............................................................................................................. 206
EXERCISE 1 ....................................................................................................... 208
POST-TEST
210
POST-TEST SOLUTIONS
212
136
MAT1581
Mathematics I (Engineering)
MODULE 7
LEARNING UNIT 1
INTEGRATION
Reverse of
differentiation I
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  ax n dx
CONTENTS
PAGE
1.
2.
3.
3.1
3.2
3.3
THE PROCESS OF INTEGRATION ................................................................. 138
THE GENERAL SOLUTION OF INTEGRALS OF THE FORM ax n ............. 139
RESPONSES TO ACTIVITIES.......................................................................... 141
Activity 1 ............................................................................................................. 141
Activity 2 ............................................................................................................. 142
Activity 3 ............................................................................................................. 143
MAT1581
Mathematics I (Engineering)
137
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
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:
If f  x   x3 , then f '  x   3x 2 . 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  3 x 2 is x 3 .
In differentiation, the differential coefficient dy
indicates that a function of x is being
dx
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  3x 2 dx means "the integral of 3x2 with respect to x”.
There are other functions that we can differentiate to obtain 3x2, for example
x 3  5, x 3  0, 5 and x 3  2. In fact, any function of the form x 3  C , where C is a
constant, will be the answer to the question find  3x 2 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  3x 2 dx  x 3  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.
138
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
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 31 cos3x .
b)
Find the indefinite integral of sin 3x.
Write your answer in the correct mathematical notation.
3. Consider the indefinite integral   3t 2  2t  3 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 ax n
The general solution of integrals of the form  ax n dx , where a and n are constants is given by
n
 ax dx 
ax n 1
C
n 1
This rule is true when n is fractional, zero, or a positive or negative integer, with the
exception of n = 1.
Example 1
1
 2 x dx   2 x dx 
a)

2 x11
11
2x2
2
C
C
 x2  C
b)
 3dx   3x dx
Remember definition a 0  1
0
Thus 1 can be written as x 0
Use x as x is the independent variable
x 01
C
1
 3x  C
3
c)





4

3
dx   4 x 2 dx
3
x2
MAT1581
Mathematics I (Engineering)
x  21
C
 32  1
3
 4.
139
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
x 2
 4. 1  C
2
1
 4. 
2
1
 8 x

x

1
2
C
1
2 C
Note that the constant multiple a or the coefficient of xn may be written outside the
ax n 1
integral sign, that is  ax n dx  a  x n dx 
C.
 n  1
Thus example c) above can be written as
3
 4
x  2 1
1
 32

dx  4 x dx  4. 3
 C  8 x  2  C
 32
 2 1
 x
Remember, each result may be checked by differentiating the answer.
ACTIVITY 2
Find the integrals and check your answer:
 3 dx
b)
 2x dx
1
c) 
 2 dx
x
d)
 y dy
 x dx
f)
 x dy
a)
0.2
e)
2
0.2
Remember to check the response on page 142.
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)
Integrate 3 x 2  5 x  7  x 2  x 2  8 x 3 with respect to x :
1
  3x  5 x  7  x  x  8x dx
  3 x dx   5 x dx   7 dx   x dx   x dx   8 x dx
2
1
2
2
3
1
2
2
2
3
3x 21 5 x11
x 2 1 x 21 8 x 31


 7x  1


C
2 1 11
2  1 3  1
2 1
1
5 2
2x 2
8 x 2
 x 1 
C
x  7x 
2
3
2
3
 x3 
140
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
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.
  3x  2 x  1 dx   3x dx   2 x dx  1dx
2
b)
2
 x3 
 x2 
 3    2    1x1  C
 3
 2 
3
2
 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.
Find   3.7  2.9 x  dx
c)
  3.7 dx   2.9 x dx
2.9 x 2
C
2
 3.7 x  1.45 x 2  C
 3.7 x 
ACTIVITY 3
a)
  3t  2t  3 dt
2
b)
  2 x4 3 
  2 x  5  x 2  dx


Remember to check the response on page 143.
3.
RESPONSES TO ACTIVITIES
3.1
Activity 1
d 2
 x   2x
dx
b) Indefinite integration reverses the process of differentiation, so we can write
2
 2x dx  x  C
1a)
2a)
b)
From our knowledge of differentiation
We always include the additional constant of integration when finding indefinite
integrals.
From our knowledge of differentiation
d
  13 cos 3x    13  3 ( sin 3x)  sin 3x
dx
Indefinite integration reverses the process of differentiation, so we can write
 sin 3x dx   13 cos 3x  C
3a)
The independent variable is t. We can tell this by inspecting the term dt.
b) The integrand is 3t 2  2t  3 .
MAT1581
Mathematics I (Engineering)
141
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
3.2
Activity 2
a)
 3 dx   3 x dx
 3  1 dx
0
Remember 3 is a constant
Constants can be taken out of the integral
 3  x 0 dx
The independent variable is x
We therefore write 1 as x 0
x 01
 3
C
0 1
 3 xC
d
dx
Check:
b)
c)
 3 x  C  3
2 x 2 1
 2 x dx  2  1  C
2 x3

C
3
2
Check:
 1 dx  x 2 dx
 2

x
x -21
=
C
2  1
x 1

C
1
1
  C
x
Check:
d)
 2.3x 2
d  2 x3
C

 2 x2


3
dx  3

Use rules for exponents
d  1 d
1
1
2
2
     1.x   1.  1x  x  2
dx  x  dx
x

1
y dy   y 2 dy


y2
C
1 1
2

d  2y2

Check:
C

dx  3


3
1 1
y2
3
3
2
C
3
 y2
2y2

C
3
142
3 1
3 2y2
 .
2 3
1

y
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 1
INTEGRATION: Reverse of differentiation I
x0.21
C
0.2  1
x1.2

C
1.2
0.2
 x dx 
e)
Check:

d  x1.2
C

dx  1.2

 1.2 
x1.21
1.2
 x 0.2
 x dy  x  1 dy
0.2
f)
0.2
 x 0.2  y 0 dy
The independent variable is y,
therefore x 0,2 must be considered as a constant
y 01
C
0 1
 x 0.2 y  C
 x 0.2
Check:
d 0.2
x y  C   x 0.2

dy
3.3
Activity 3
a)
  3t  2t  3 dt
2
 t 3  t 2  3t  C
b)
Check:
d 3 2
 t  t  3t  C 
dt
 3t 2  2t  3
  2 x4 3 
  2 x  5  x 2  dx


2 x3 1 x5
x 21

 . 3
C
2  1
3
5 5
2 x3 x 5 3 x 1



C
3
25 1
2 x3 x5 3


 C
3
25 x
Check:

d  2 x3 x5 3

 C

25 x
dx  3

2.3.x 2 5 x 4

 (1)3 x 2
3
25
4
3
x
 2 x2 
 2
5 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  ax n dx
We further investigate the reverse of differentiation in learning unit 2.
MAT1581
Mathematics I (Engineering)
143
MODULE 7
LEARNING UNIT 2
INTEGRATION
Reverse of
differentiation II
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
CONTENTS
PAGE
1.
2.
3.
4.
4.1
4.2
4.3
SIMPLIFY THE INTEGRAND BEFORE INTEGRATION ............................. 145
DETERMINE THE VALUE OF THE CONSTANT OF INTEGRATION ....... 147
APPLICATION: DISPLACEMENT, VELOCITY, ACCELERATION ............ 148
RESPONSES TO ACTIVITIES.......................................................................... 149
Activity 1 ............................................................................................................. 149
Activity 2 ............................................................................................................. 150
Activity 3 ............................................................................................................. 150
MAT1581
Mathematics I (Engineering)
144
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
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)
Find  x 1  x 2  dx
2
The integrand is not in the general form axn and looks complicated. We know how
to expand 1  x 2  and then multiply the answer by
2
1
x  x 2 . So let’s apply our
knowledge and hope that we find an integral that we know how to solve.
Simplify the integrand:
1  x   1  2 x  x
 x 1  x   x 1  2 x  x 
2 2
2
2 2
1
4
2
2
4
 x 2  2x 2  x 2
5
1
9
All the above terms are of the form ax n
Now we are able to find the integrand using the general solution from integration
in unit 1.


  x 1  x 2    x 2  2 x 2  x 2 dx
2
3

5
1
7
x2

3
2
3
9
11
2x 2
x 2
 11  C
7
2
2
7
11
2x
4x
2x 2



C
3
7
11
2
2

d  2x 2 4x 2 2x 2
Check:


C

7
11
dx  3

3
7
11
2 11 9 
2 3 1 4 7 5
 . x 2  . x 2  . x 2
7 2
11 2
3 2


 x 2  2x 2  x 2
5
1
9

Note that this check only confirms the correct value of the integral
5
9
1
 x 2  2 x 2  x 2 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.
MAT1581
Mathematics I (Engineering)
145
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
b)
 3x3  2 x 2  8 x  1
Find 
x6

dx
Start by splitting the integrand into four fractions. Divide each term of the
numerator by the denominator.
3x3  2 x 2  8 x  1 3x3 2 x 2 8 x 1
That is
 6  6  6  6
x6
x
x
x
x
Now simplify each fraction
3
2
8
1
 3 4  5 6
x
x
x
x
Use rules for exponents
 3 x 3  2 x 4  8 x 5  x 6
 3x3  2 x 2  8 x  1
Thus

x6

  3x  2 x  8 x  x  dx
3
4
5
6
3 x 2 2 x 3 8 x 4 x 5



C
2
3
4
5
3
2
1
  x 2  x 3  2 x 4  x 5  C
2
3
5

2
c)
 x 3
Find 


1  7 x 
2
dx
3x3
We can simplify the integrand under the integral sign.
2
2
 x 3 1  7 x 

Thus 
dx

3x3

1  14 x  49 x  dx
2
2
 x 3
 


3x3
 x 3  14 x 3  49 x 3
 
dx
3x3

5
2
8
14 5
49 8 3 3 
1 2
   x 3  3  x 3 3 
x  dx
3
3

3
7
4
1
  3
14 x  3 49 x  3 
x
  


 dx

3
3 
 3
7
4
1
x 3
 14 x 3
 49 x 3
dx  
dx  
dx

 3
 3
 3
x  3 1
14 x  3 1
49 x  3 1



C
3   73  1 3   43  1 3   13  1
7
4
1
49 x 3
x 3
1

 14 x  3 
C
4
2
4
146
2
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
ACTIVITY 1
Determine
3
a)  1  u  du
b)
 x  x  1 x  2  dx
c)
 x3  6 x 2  5 x

dx

x

Remember to check the response on page 149.
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
If u    3 x 2  2 x  1 dx, find the function u if u  10 when x  1 .
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.
  3x  2 x  1 dx  x  x  x  C
2
3
2
Thus u  x3  x 2  x  C
(*)
Given u  10 if x  1
Substitute in  *
10  1  1  1  C
now C can be determined
C  7
Answer: u  x3  x 2  x  7
ACTIVITY 2
1. If y    3 x 2  x  10  dx, find the function y if y  0 when x  2


2. If y   4 x3  3x 2  6 x  2 dx, determine the value of y when x  4,
given that when x  2, y  30
Remember to check the response on page 149.
MAT1581
Mathematics I (Engineering)
147
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
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 v 
ds
.
dt
The displacement s is therefore given by s   v dt .
The acceleration a is the derivative of the velocity v, that is a 
dv
.
dt
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
v  4t  t 2 in metres per second. What is the change in displacement in the first 4
seconds?
As v 
ds
, we have s   v dt.
dt


Thus s   4t  t 2 dt
4t 2 t 3

 C
2
3
To find the change in displacement we need the displacement at time t = 0 and t = 4.
4(0) 2 (0)3
At t  0,. s0 

C
2
3
 00C
C
4(4) 2 (4)3

C
2
3
192  128

C
6
64

C
6
Now the change in displacement s
At t  4,. s4 
 s4  s0
64
C C
6
 10, 6 metres

148
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
ACTIVITY 3

1 
1. An object is moving in a straight line with a velocity v   t 
 m/s .
t

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.
RESPONSES TO ACTIVITIES
4.1
a)
Activity 1
Simplify the integrand. The binomial theorem should be used, or write
1  u   1  u 1  u 1  u 
3
1  u   1  3u  3u 2  u 3
3
All the above terms are of the form ax n
  1  u    1  3u  3u 2  u 3  du
3
b)
3
u4
 u  u 2  u3   C
2
4
 x  x  1 x  2  dx
  x  x 2  3x  2  dx
   x3  3x 2  2 x  dx
x 4 3x3 2 x 2


C
4
3
2
x4

 x3  x 2  C
4

c)
 x3  6 x 2  5 x
  x3 6 x 2 5 x 

dx    
  dx

x
x
x 

 x
   x 2  6 x  5  dx

x3
 3x 2  5 x  C
3
4.2
Activity 2
1.
x2
  3x  x  10  dx  x  2  10 x  C
2
MAT1581
Mathematics I (Engineering)
3
149
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
x2
Thus y  x 
 10 x  C (*)
2
Substitute in * given y  0 when x  2
3
 0   2 
 2
2
 10  2   C
2
0  8  2  20  C
3
now C can be determined
 C  10
x2
 10 x  10
Answer: y  x 
2
3
2.


y   4 x 3  3 x 2  6 x  2 dx,
4x 4 3x3 6 x 2


 2x  C
4
3
2
 x 4  x3  3x 2  2 x  C
given that when x  2, y  30 we find the value of C
=
y  x 4  x3  3x 2  2 x  C
30  (2)4   2   3  2   2  2   C
3
2
30  16  8  12  4  C
C  30  16
 14
Thus y  x 4  x3  3x 2  2 x  14
and the value of y when x  4
y   4   (4)3  3(4) 2  2  4   14
4
 256  64
 48
8
 14
 246
4.3
1a)
Activity 3
Position refers to displacement.
At the origin implies that at the start v = 0, t = 0 and s = 0.
1

1 
t2 t 2
s   t 
dt

 1 C

2
t

2
At the origin s  0, t  0 thus in the above equation 0  0  0  C and C  0
150
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 2
INTEGRATION: Reverse of differentiation II
1
At t  15,
b)
t 2 2t 2
s  
2
1
2
(15)

 2 15
2
 120.3 m
Given the velocity, you have to find the acceleration, that is differentiation.
1
 
dv d 
1  d
2
a 
t
t
 t 





dt dt 
t  dt 

 1 3 
 1   t 2 
 2



1
 1
2 t3
At t  15,
a  1
1
2 (15)3
 0.99 m/s 2
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
v  a  dt
integral.
 a  1 dt
 a  t 0 dt
 at  C
Thus v  at  C where C is the constant of integration.
b)
Evaluate the equation in a) with t = 0 and v = u.
u  a (0)  C
u C
Thus v  at  u
or in the more usual form
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.
MAT1581
Mathematics I (Engineering)
151
MODULE 7
LEARNING UNIT 3
INTEGRATION
Method of substitution
OUTCOME
At the end of this learning unit, you should be able to determine integrals of the form
 f '  x   f  x   dx using the substitution u = f(x).
n
CONTENTS
PAGE
1.
2.
METHOD OF SUBSTITUTION ........................................................................ 153
RESPONSE TO ACTIVITY ............................................................................... 156
MAT1581
Mathematics I (Engineering)
152
Module 7 Learning unit 3
INTEGRATION: Method of substitution
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   3 x  7  dx,  x 4  6 4 x3 dx, or  3 4x3  8  x 2 dx. The
10
3
first integral could be found by expanding  3 x  7  and then integrating, but this would
10
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   3 x  7  dx .
10
In this integral we will use the substitution
Differentiating with respect to x we get
u
du
dx
 du
  3x  7 
3
 3dx
du
and dx 
3
Now substitute in the original integral
10 du
  3x  7  dx   u
10
3
Let u   3x  7  and du  3 dx
1 10
u du  Remember constants can be taken out 
3
After substitution the new integrand should only contain constants and the new

independent variable u. The new integral is of the form ax n .
1
10
Thus   3x  7  dx   u10 du
3
1 u11
 
C
3 11
u11

C
33
Substituting for u we obtain the answer in the original variable x
11
3x  7 


C
33
MAT1581
Mathematics I (Engineering)
153
Module 7 Learning Unit 3
INTEGRATION: Method of substitution
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
f  x  dx as u n du with the possible exception of a multiplicative constant.
A more general form of the basic rule is
u n 1
n

 C , n  1
u
du

n 1
Example 1
a)


3
Determine  x 4  6 4 x 3 dx
In this integral we will use the substitution
Differentiating with respect to x we get
u
 x4  6
du
dx
 du
 4 x3
 4 x3dx
and dx 
du
4 x3
Now substitute in the original integral
  x  6  4 x dx   u du
4
3
3
Let u  x 4  6 and du  4 x3dx
3
u4
C
4
Substituting for u we obtain the answer in the original variable x

 x  6  C

4
4
4
b)
Determine
Let
 3 4x  8  x dx .
3
2
u  4 x 3  8,
du
 12 x 2
dx
du
and dx 
12 x 2
du
or x 2 dx 
12
then
Thus  3 4x 3  8  x 2 dx


1
3
2

 4 x  8 3 .x dx

1
du

  u3.
12

154
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 3
INTEGRATION: Method of substitution
4
1 u3
 . 4 C
12 3



4
3
3u
C
48
4
u3
16
C
 4x
3
4
8 3
16
 C
 4 x  8  C

3
3
4
16
c)
1

Find 
dx
2
 x 3
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)


2
Evaluate  x3  5 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.
  x  5 dx    x  10 x  25 dx
3
2
6
3
x 7 10 x 4

 25 x  C
7
4
x7 5x 4


 25 x  C
7
2

Using the above examples, we can write a new general solution in f(x) notation as
follows:
n 1
f(x)

 c, n  1
  f(x) .f '(x) dx 
n
n 1
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.
MAT1581
Mathematics I (Engineering)
155
Module 7 Learning Unit 3
INTEGRATION: Method of substitution
ACTIVITY 1
Determine the following integrals:
c)
  x  2  2 x dx
 3 x  1 dx
e)

2
2
a)
 x  3x  5  2 x  3 dx
  x  2  x dx
2
b)
25  x 2 dx
3
2
d)
f)




dt
t 1
g)

1

dx
2

 60 x  100  9 x
h)
 1  2x dx
i)
 x 1  x dx
j)
 1  x  dx
l)
  14 1  x  4 dx
n)
 3x 2
dx


3 2
 1  x 
p)
2
3
 (1  3x )  4  x  x  dx
2

3x 2  6
k) 
dx
 2 x 3  12 x  10


3
7

3
m)  x 1  x 2 2 dx
o)
2
3
 15 x 1  x  dx
q)

dx

2
 1  2 x 
4
2.
RESPONSE TO ACTIVITY
a)
Let I    x 2  2  2 x dx and let
2
and
I
2
u  x2  2
Then
Thus our integral becomes
3
du
 2x
dx
du  2 x dx
  u 2 du

u3
C
3
 x  2  C

3
2
3
b)
I   x 2  3 x  5  2 x  3 dx
Let
u  x 2 +3 x +5
Then
du   2 x  3 dx
Therefore I   u du   u 2 du
1
3
2u 2

C
3

156
2  x 2  3x  5
3
3
2
C
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 3
INTEGRATION: Method of substitution
c)
Let I   3 x  1 dx and let 3x  1  u
3
Then
du
 differentiate with respect to x on both sides 
dx

1 
 I   u  du 
3 

1 1
  u 2 du
3
1 2 3
 . u 2 C
3 3
3
2
  3x  1 2  C
9
d)
Let I    x 2  2  x dx and let u  x 2  2
3
du
 2x
dx
and du  2 x dx
Then
3
1
x 2  2   2 x  dx


2
1
  u 3 du
2
1 u4

C
2 4
I 
 x  2  C

4
2
8
e)
Let I
  25  x 2 dx and
u  25  x 2
du
 2 x
dx
du  2 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.
then
f)

 
dt
and t  1  u , then dt  du
t 1
 du
1
Thus I   1   u  2 du
2
u
1
 2u 2  C
Let I

 2  t  1 2  C
1
MAT1581
Mathematics I (Engineering)
157
Module 7 Learning Unit 3
INTEGRATION: Method of substitution
g)


1
dx

dx   2
2

 60 x  100  9 x
 9 x  60 x  100

dx


 3 x  10 
  
Let I
2

1
dx
2
 60 x  100  9 x
 
Rearrange the denominator
Factorise the denominator
and u  3 x  10
then du  3dx
Thus I   13  u du
2
  13 .  11  u 1  C

h)
 3x  10 
1
C
3
 1  2x dx
3
Let u  1  2 x, then du  2 dx or 12 du  dx
1
Thus  3 1  2 x dx  12  u 3 du
4
  12   34  u 3  C
4
 83 1  2 x  3  C
 83 3 1  2 x   C
4
i)
 x 1  x dx
2
Let u  1  x 2 , then du  2 x dx or 12 du  x dx
1
Thus  x 1  x 2 dx  12  u 2 du
3
  12   23  u 2  C
3
 13 1  x 2  2  C
1  x   C
2 3
 13
j)
 1  x  dx
7
Let u  1  x, then du  dx
Thus  1  x  dx   u du
7
7
 18 u 8  C
 18 1  x   C
8
158
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 3
INTEGRATION: Method of substitution
k)

3x 2  6
dx

 2 x 3  12 x  10
Let u  2 x 3  12 x  10, then du  (6 x 2  12) dx
thus 12 du  (3 x 2  6) dx

3x 2  6
Thus 
dx
 2 x 3  12 x  10
 du 1  2
 
 2  u du
 u
1
1
2
1
  12  12  u 2  C
1
2
  2 x  12 x  10   C
3
 2 x3  12 x  10  C
l)
3
  14 1  x  4 dx

Let u  1  x, then du   dx
Thus  
1
4
1  x 

3
4
dx 
1
4
u

3
4
du
1
4
    u  C
1
4
4
1
1
 1  x  4  C
3
m)
2
 x 1  x  2 dx
Let u  1  x 2 , then du  2 x dx or 12 du  xdx
3
2
Thus  x 1  x  dx 
2
1
2
3
2
 u du
5
  12   52  u 2  C
5

n)
1  x 2  2
5
C
 3x 2
dx


3 2
 1  x 
Let u  1  x3 , then du  3x 2 dx
 3x 2
dx
Thus 

3 2
 1  x 
  u 2 du
 (1) u 1  C
1

C
1  x3 
MAT1581
Mathematics I (Engineering)
159
Module 7 Learning Unit 3
INTEGRATION: Method of substitution
o)
 15 x 1  x  dx
3 4
2
Let u  1  x3 , then du  3 x 2 dx
Thus  15 x 2 1  x3  dx  5 3x 2 1  x3  dx
4
4
4
 5 u du
 5  15  u 5  C
 1  x3   C
5
p)
 (1  3x )  4  x  x  dx
3 2
2
Let u  4  x  x 3 , then du  (1  3 x 2 ) dx
Thus  (1  3 x 2 )  4  x  x 3  dx
2
  u 2 du
  1 u 1  C
   4  x  x3   C
1

q)
1
C
(4  x  x 3 )

dx

2
 1  2 x 
Let u  1  2 x, then du  2 dx or 12 du  dx

dx
Thus 
2
 1  2 x 
 12  u 2 du
  12  (1) u 1  C

1
C
2 1  2 x 
This the end of learning unit 3 and you should therefore be able to determine integrals of
the form  f '  x   f  x   dx using the substitution u = f(x).
n
Now we move on to learning unit 4 with a focus on standard integrals.
160
MAT1581
Mathematics I (Engineering)
MODULE 7
LEARNING UNIT 4
INTEGRATION
Standard integrals
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
CONTENTS
1.
2.
PAGE
TABLE OF STANDARD INTEGRALS ............................................................ 162
 f (x)
INTEGRALS OF THE FORM 
dx .......................................................... 163
 f(x)
3.
INTEGRALS OF THE FORM  f '  x  ef  x  dx .................................................. 165
4.
INTEGRALS OF THE FORM  f (x).af(x) dx .................................................. 166
5.
5.1
5.2
5.3
5.4
RESPONSES TO ACTIVITIES.......................................................................... 168
Activity 1 ............................................................................................................. 168
Activity 2 ............................................................................................................. 168
Activity 3 ............................................................................................................. 169
Activity 4 ............................................................................................................. 169
MAT1581
Mathematics I (Engineering)
161
Module 7 Learning unit 4
INTEGRATION: Standard integrals
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
a x(n 1 )

 c,
n 1
1.
 ax dx
2.
  f(x) .f'(x) dx
3.
 f(x)
4.
 f (x).e
5.

6.
 f (x).sin f(x) dx
  cos f(x)  c
7.
 f (x).cos f(x) dx
 sin f ( x)  c
8.
 f (x). tan f(x) dx
 n sec f(x)  c  n cos f(x)  c
9.
 f (x).cot f(x) dx
 n sin f(x)  c
10.
 f (x).sec f(x) dx
 n sec f(x)  tan f(x)  c
11.
 f (x).cosec f(x) dx
 n cosec f(x)  cot f(x)  c
12.
 f (x).sec f(x) dx
 tan f(x)  c
13.
 f (x).cosec f(x) dx
  cot f(x)  c
14.
 f (x).sec f(x). tan f(x) dx  sec f(x)  c
15.
 f (x).cosec f(x).cot f(x)dx  cosec f(x)  c
162
n
n
f (x)
n 1
f(x)


 c, n  1
n 1
 n f(x)  c
dx
f(x)
n  1
dx
f (x).a f(x) dx
2
2
 e f(x)  c

a f(x)
c
n a
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 4
INTEGRATION: Standard integrals
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.
2.
 f (x)
dx
INTEGRALS OF THE FORM 
 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.


dy 1
1
1

thus 
 dx   x dx  n x . We can therefore
x
dx x
see number 3 in the table as the special case where n = .
Remember: If y  n x, then
In general,
d
f '( x)
 f '( x)
thus 
dx  n f ( x) .
n f  x   

dx
f ( x)
 f ( x)
Example 1
a)
1  2x
dx
Determine 

 7  x  x2
du
Put u  7  x  x 2 , then
 1  2 x or du  1  2 x  dx
dx
1  2x
 1 du
Thus 
dx



u
 7  x  x2
 n u  C
 n 7  x  x 2  C
Some of you might recognise immediately that
the derivative of 7  x  x 2 is 1  2 x.
f '( x)
Therefore the integrand is of the form
f ( x)
and the answer can be read from the table as
n 7  x  x 2  C
MAT1581
Mathematics I (Engineering)
163
Module 7 Learning unit 4
INTEGRATION: Standard integrals
b)

1
dx
Determine 
x
1

x



This is more difficult to recognise, so we need to do a substitution.

1

du 1  2
then
 x
dx 2
Put u  1  x ,
and
1

1
du  2 x 2 dx
1 dx
 . 1
2 2
x
dx
2 du 
x

1
dx

1
x
x





1
dx
.

x
 1 x


1

 2 du
u
1
 2
 du
u
Now it is possible to recognise that the integrand is of the form
f ' x
f  x
 2 n u  C
 2 n 1  x  C
ACTIVITY 1
Determine
a)  u 1 du
 3x
dx
b) 
 1  x4
3
2
 x  3x  3x  1
dx
c) 
x2

Remember to check the response on page 168.
3
164
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 4
INTEGRATION: Standard integrals
3.
INTEGRALS OF THE FORM  f '  x  ef  x  dx
dy
 e x from which it follows that  e x dx  e x  C
dx
In general the standard integral is obtained as follows:
dy
f x
f x
f x
f x
 f '  x  e   , thus  f '  x  e   dx  e    C
If y  e   , then
dx
If y  e x , then
Example 2
a)
Find  e3 x dx.
The most direct route would be to say, let f  x   3 x then f '  x   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 equality.
 e dx   (3)e dx   3e dx
3x
3x
1
3
3x
1
3
The integrand is of the form f '  x  e   .
f x
Use the table of standard integrals to read off the answer.
  e3 x dx  13 e3 x  C.
Or we may use substitution.
du
3
dx
 du  3dx or
Let u  3 x, then
dx  13 du
Hence,
 u
1
 e dx   e . 3 du
3x
 13  eu du
 13 eu  C
 13 e3 x  C
b) Find  e x .2 x dx .
2
If f  x   x 2 , then f '  x   2 x, the integrand is of the form f '  x  e f  x 
Use the table of standard integrals to write down the answer.
x
x
 e .2 x dx  e  C
2
2
MAT1581
Mathematics I (Engineering)
165
Module 7 Learning unit 4
INTEGRATION: Standard integrals
Or use substitution:
Let u  x 2 , then du  2 x dx
Hence
x
u
 e .2 x dx   e du
2
 eu  C
 ex  C
2
1
Determine 
 3 x dx .
 7e
We will find the answer without substitution.
 1 dx  1 e 3 x dx
 3x
7 
 7e
c)
 17  31  3 e 3 x dx
  211  3e 3 x dx
  211 e 3 x  C
ACTIVITY 2
Determine
a)   e 7 x  3e5 x  11e9,2 x  5e 2 x  dx
b)
1 x x  1  2 x dx
e
2


Remember to check the response on page 168.
INTEGRALS OF THE FORM  f (x).af(x) dx
4.
If y  a x , then
dy
 a x n a from which it follows that  a x n a dx  a x  C.
dx
Sometimes we would slightly modify the expression to give a neater standard integral.
In this case n a is a constant, which can therefore be taken out of the integral
n a  a x dx  a x  C
Dividing both sides by n a :
ax
C
 a dx  n a  n a where the last term can be replaced with a single constant symbol.
x
In general the standard integral is

166
f (x).a f(x) dx 
a f(x)
C
n a
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 4
INTEGRATION: Standard integrals
Example 3
a)
Find  a 3 x dx
Let u  3x, then
du
 3 or dx  13 du
dx
  a 3 x du  13  a u du
 13 .

b)
au
C
n a
a3 x
C
3 n a
Determine  x.7 3 x 1 dx .
2
Put u  3 x 2  1, then
du
du
 6 x and
 x dx
dx
6
Thus  x.73 x 1 dx  16  7u du
2
 16
7u
C
n 7
73 x 1

C
6 n 7
2
ACTIVITY 3
Determine
a)  a 2 x dx
b)
 3 dx
5x
Remember to check the response on page 169.
To consolidate what you have learnt thus far, try an activity with mixed questions.
ACTIVITY 4
Determine
x2
2 1 
 x2
dx
a) 
  xe  xa  7 x 1  x




1


3
b)  e x  1 e x dx
3
 4x
dx
e) 
 4  x4
 x  2x  1
dx
d) 
x3

1  2x
dx
f) 
 2
 x  x 1
g)  1  x  x dx
h)   3s  4  ds
 ex
c)  2 dx
 x
i)
2
2
 3x 1  2 x dx
2
Remember to check the response on page 169.
MAT1581
Mathematics I (Engineering)
167
Module 7 Learning unit 4
INTEGRATION: Standard integrals
5.
RESPONSES TO ACTIVITIES
5.1
Activity 1
a)
 u du   u du  nu  C
1
1
b)
 3x
Determine 
dx
4
 1 x
du
du
 4 x3 , thus
 x3dx
Put u  1  x 4 , then
4
dx
Substitute values then
3
 3x
dx

4
 1 x
3
1 du
 3

u 4
3 1
 
 du
4 u
 34 n u  C
 34 n 1  x 4  C
c)
First simplify the integrand:
  x3 3x 2 3x 1 
   2  2  2  2  dx
x
x
x 
 x
3


   x  3   x 2  dx
x


You should be able to do this without any substitution
 x  3x  3x  1
dx

x2

3
2
x2
 3 x  3 n x  x 1  C
2
x2
1

 3 x  3 n x   C
x
2

5.2
Activity 2
a)
  e  3e  11e
168
7x
5x
9,2 x
 5e 2 x  dx

e7 x 3e5 x 11e9,2 x 5e 2 x



C
7
5
9, 2
2

e7 x 3 5 x 11 9,2 x 5 2 x
 e 
e  e C
7 5
9, 2
2
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 4
INTEGRATION: Standard integrals
b)
Determine  e
1 x  x  1  2 x dx

2
Put u  1  x  x 2 , then
Thus  e

du
 1  2 x and du  1  2 x  dx
dx
1 x  x  1  2 x dx  eu du



2
 eu  C
e
5.3
Activity 3
a)
 a dx  12  2 a dx
2x
2
2x

5.4
1 x  x   C
b)
a2x
C
2 n a
 3 dx  15  5.3 dx
5x
5x

35 x
C
5 n 3
Activity 4


  xe x2  xa x 2  7 x 1  x 2 1  dx



du
du
 2 x, and
 xdx
Put u  x 2 , then
dx
2
1
 x2  xa x 2
Thus 
 7 x 1  x 2  dx
  xe


a)


 12  eu du  12  a u du  72  (1  u ) 1 du
 12 eu
 12
au
n a
2
 12 e x
b)
2
ax

2 n a
 72 n 1  u 

 72 n 1  x 2
C

C
  e  1 e dx
x
3 x
Put u  e x  1,
du
then
 ex
dx
and du  e x dx
MAT1581
Mathematics I (Engineering)
169
Module 7 Learning unit 4
INTEGRATION: Standard integrals
Thus
  e  1 e dx   u du
3 x
x
3

u4
C
4
 e  1  C

4
x
4
1
c)
 ex
 2 dx
 x
Put u  x 1 , then
du
dx
 1 x 2 and  du  x 2 dx  2
dx
x
1
Thus
 ex
 2 dx
 x
   eu du
  eu  C
1
 e x  C
d)
2
1
1 2
 x  2x  1
dx     2  3  dx

3
x
x 

 x x
1

    2 x 2  x 3  dx

 x
2 x 1 x 2

C
1
2
2
1
 n x   2  C
x 2x
 n x 
e)
3
 4x
dx

 4  x4
The integrand is of the form
f ' x
f  x
 n 4  x 4  C
f)
 1  2 x dx
 2
 x  x 1
The integrand is of the form
f ' x
f  x
 n x 2  x  1  C
g)
170
 1  x  x dx
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 4
INTEGRATION: Standard integrals
  x dx   x x dx
1
3
  x 2 dx   x 2 dx
3
[using the law a m .a n  a m  n ]
5
2
2
   x2    x2  C
3
5
h)
  3s  4  ds  13  3  3s  4  ds
2
2
 13

 3s  4 
3
C
3
 3s  4 
3
C
9
or
  3s  4  ds    9s  24s  16  ds
2
2
9 s 3 24s 2

 16s  C
3
2
 3s 3  12s 2  16 s  C

The two answers should be the same.
To compare expand the fraction in answer one using the binomial theorem:
 3s  4 3
9
i)
(3s )(4) 2  43
 3s 3  3(3s)2 (4)  3(2)
2
C
9
27 s 3  108s 2  144s  64

C
9
 3s 3  12s 2  16s  C , where 649 is incorporated in C



1
2

 3x 1  2 x dx  3 x 1  2 x 2 dx
2
1
2 2
3
  4  4 x 1  2 x
dx



 34
  12

3
2 2
1  2x
  
2
3

1  2 x   C
2 3
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.
MAT1581
Mathematics I (Engineering)
171
MODULE 7
LEARNING UNIT 5
INTEGRATION
Partial fractions
OUTCOME
At the end of this learning unit, you should be able to use partial fractions to integrate
algebraic expressions.
CONTENTS
PAGE
1.
2.
INTEGRATION USING PARTIAL FRACTIONS ............................................ 173
RESPONSE TO ACTIVITY ............................................................................... 175
MAT1581
Mathematics I (Engineering)
172
Module 7 Learning unit 5
INTEGRATION: Partial fractions
1.
INTEGRATION USING PARTIAL FRACTIONS
The first step is to verify that you do not have an integral of one of the following forms:
1.
EXAMPLE
 2x  2
5
2
dx

x

2
x
 2 x  2  dx


 2
5

  x  2x
  f  x   f '  x  dx, n  1
n
 x  2x  C

4
2
4
2.
 f ' x
 f x dx
  
 2x  2
dx  n x 2  2 x  C
 2
  x  2x
Partial fractions may be used in integration of algebraic expressions as follows:
Example 1
3x  8
dx with the aid of partial fractions.
a) Find the integral of 
 2
 x  2 x  35
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.
That is,
3x  8
13
23


x  2 x  35 12  x  7  12  x  5
2
This gives us two integrals of form 2.
13
23


 3x  8
dx  
dx 
dx
 2
 x  2 x  35
 12  x  7 
 12  x  5 
13
n x  7
 12
23
n x  5
 12

C
 3x 2  7
b) Use partial fractions to find 
dx
4
  x  1
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.
3x 2  7
 x  1
4

3
 x  1
2

6
 x  1
3

4
 x  1
4
This gives us three integrals of form 1.
MAT1581
Mathematics I (Engineering)
173
Module 7 Learning unit 5
INTEGRATION: Partial fractions
 3x 2  7

 6

3
4
dx  
dx 
dx  
dx

4
2
3
4
  x  1
  x  1
  x  1
  x  1
c)
Find 



3
 x  1


3
 x  1

6
2  x  1
3
 x  1
2
2


4
3  x  1
3
 C
3
 C
4
3  x  1
x3  2 x 2  4 x  4
dx .
x2  x  2
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.
x3  2 x 2  4 x  4
4
3
 x3

2
x  x2
 x  2   x  1
This gives four integrals
3
2
 x  2 x  4 x  4 dx  x dx  3 dx   4 dx   3 dx





x2  x  2

  x  2
  x  1
x2
 3 x  4 n x  2 3 n x  1  C
2
using the rules of logarithms we can combine
the two terms with logarithms

x2
2
 3 x  n  x  2 
x2

2
 x  2
 3 x  n
3
 x  1

4
 n ( x  1)3  C
4
C
ACTIVITY 1
Find

2x  8

a)
 2
 x  5x  6
 x3  3x 2  7 x  4

b)
dx

x2  2x

c)
 8x2
dx

3
 3
  x  2
 4 x  8 dx
 2
 x  4x  5
Remember to check the response on page 175.
d)
174
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 5
INTEGRATION: Partial fractions
2.
RESPONSE TO ACTIVITY
a)
First resolve the integrand in partial fractions:
2x  8
2x  8
A
B



2
x  5x  6
 x  3 x  2   x  3  x  2 
Combine right-hand side:
A  x  2   B  x  3
2x  8

 x  3 x  2 
 x  3 x  2 
Equating numerators gives:
2 x  8  A  x  2   B  x  3
Let x  2, then 2(-2)  8  B(2  3)
4 B
Let x  3, then 2(3)  8  A(3  2)
2  A
A  2
4
2
2x  8


 x  3  x  2   x  3  x  2 
The integral becomes
 2x  8
 2
 x  5x  6
  2
4 
  

 dx
   x  3  x  2  
 dx
 dx
 2
 4
  x  3
  x  2
 2n x  3  4n x  2  C
using the rules of logarithms we can simplify our answer
by combining the two logarithm terms
  n  x  3 
 n  x  2 
 n
MAT1581
Mathematics I (Engineering)
4
 x  2
2
 n  x  2 
4
 C
 n  x  3   C
2
4
( x  3) 2
C
175
Module 7 Learning unit 5
INTEGRATION: Partial fractions
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:
x 1
x 2  2 x x3  3x 2  7 x  4
x3  2 x 2
x2  7 x
x2  2x
5x  4
x3  3x 2  7 x  4
5x  4
 x 1 2
Thus
2
x  2x
x  2x
5x  4
in partial fractions
x2  2x
5x  4
5x  4
A
B

 
2
x  2x x  x  2 x x  2
Resolve
Combine right-hand side:
A  x  2   Bx
5x  4

x  x  2
x  x  2
Equating numerators gives:
5 x  4  A  x  2   Bx
Let x  2, then 5  2   4  2 B
 2 B  6
B3
Let x  0, then 4  2 A
A2
Therefore
176
5x  4 2
3
 
2
x  2x x x  2
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 5
INTEGRATION: Partial fractions
The integral becomes

 x3  3x 2  7 x  4
2
3 
 x 1




 dx
2


x x2
x  2x


1
  x dx   dx 2
 x


1
dx
 x2
dx  3
2
x
x
2 n x
 3 n x  2  C
2
using the rules of logarithms we can simplify our answer
by combining the two logarithm terms
x2
3

 x  n x 2  x  2   C
2
c)
This is an example of standard form 2 with n a negative number.
 8x2
3
2
3
8
dx
3
x
x
2
dx






3
3

 3
  x  2
 8   x  2
 
C
2
3
4

C
2
3
3 x  2
3
d)
2
This is an example of standard form 3.
 4 x  8 dx  2 2 x  4 dx
 2
 2
 x  4x  5
 x  4x  5
 2 n x 2  4 x  5  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.
MAT1581
Mathematics I (Engineering)
177
MODULE 7
LEARNING UNIT 6
INTEGRATION
Trigonometric
functions
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
CONTENTS
PAGE
1.
2.
2.1
2.2
INTEGRATION OF TRIGONOMETRIC FUNCTIONS .................................. 179
RESPONSES TO ACTIVITIES.......................................................................... 182
Activity 1 ............................................................................................................. 182
Activity 2 ............................................................................................................. 184
MAT1581
Mathematics I (Engineering)
178
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
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
a x(n 1 )

 c,
n 1
1.
 ax dx
2.
  f(x) .f'(x) dx
3.
 f(x)
4.
 f (x).e
5.

6.
 f (x).sin f(x) dx
  cos f(x)  c
7.
 f (x).cos f(x) dx
 sin f ( x)  c
8.
 . tan f(x) dx
 f (x)
 n sec f(x)  c  n cos f(x)  c
9.
 f (x).cot f(x) dx
 n sin f(x)  c
10.
 f (x).sec f(x) dx
 n sec f(x)  tan f(x)  c
11.
 f (x).cosec f(x) dx
 n cosec f(x)  cot f(x)  c
12.
 f (x).sec f(x) dx
 tan f(x)  c
13.
 f (x).cosec f(x) dx
  cot f(x)  c
14.
 f (x).sec f(x). tan f(x) dx  sec f(x)  c
15.
 f (x).cosec f(x).cot f(x)dx  cosec f(x)  c
n
n

f (x)
n 1
f(x)


 c, n  1
n 1
 n f(x)  c
dx
f(x)
n  1
dx
f (x).a f(x) dx
2
2
MAT1581
Mathematics I (Engineering)
 e f(x)  c

a f(x)
c
n a
179
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
Example 1
a)
Find  sin 5x dx
du
du
 5 and
 dx
dx
5
1
Hence  sin 5 x dx   sin u . du
5
1
  cos u  C
5
1
  cos 5 x  C
5
This is an example of standard form 6.
Let u  5 x, then
b)
Find  cos  2 x  1 dx
du
du
 2 and
 dx
dx
2
1
Hence  cos  2 x  1 dx   cos u . du
2
1
 sin u  C
2
1
 sin (2 x  1)  C
2
This is an example of standard form 7.
Let u  2 x  1, then
c)
Find  sin  tan x  sec 2 x dx
Let u  tan x, then du  sec2 x dx
  sin  tan x  sec2 x dx   sin u du
  cos u  C
  cos  tan x   C
ACTIVITY 1
Find
a)  cos 7x dx
 tan  5 x  7  dx
c)  x cot x dx
d)  sec  3 x  4  tan  3 x  4  dx
e)  x 2 sec 2 x 3 dx
f)  x cosec 2 x 2 dx
b)
2
Remember to check the response on page 182.
180
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
Example 2
a)
Find  e tan x .sec 2 x
Let u  tan x, then du  sec 2 x dx
  e tan x .sec 2 x dx   eu du
 eu  C
 e tan x  C
b)
sin x  cos x
Find 
dx


cos x
First simplify the integrand
 sin x  cos x dx    sin x  cos x  dx




cos x
  cos x cos x 
   tan x  1 dx
  tan x dx   1 dx
We can now use the table of standard integrals.
=n sec x  x  C
b)
Evaluate   tan 2 x  1 dx .
2
First expand the integrand as we cannot recognise a standard form.
  tan 2 x  1 dx    tan 2 x  2 tan 2 x  1 dx
   tan 2 2 x  1  2 tan 2 x  dx


   sec 2 2 x   2 tan 2 x  dx


2
  sec 2 x dx   2 tan 2 x dx
2
2
du
 dx
2
Substitute these values in the above integrals:
Now let u  2 x, then du  2 dx and
1
  sec 2 u du   tan u du
2
1
 tan u  n sec u  C
2
1
 tan 2 x  n sec 2 x  C
2
MAT1581
Mathematics I (Engineering)
181
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
ACTIVITY 2
Determine
a)  a sin x cos x dx
 e sin 2 x dx
c)  e cos e dx
d)  tan x dx
b)
3 cos 2 x
x
x
2
1
dx
e) 

 cos 2 x
f)  cos 2 x sin x dx
 sec 3x tan 3x dx
h)  sin  2x  dx
i)  tan 3 x sec 2 3 x dx
g)
4
 1  cos x dx

 sin x
Remember to check the response on page 184.
j)
2.
RESPONSES TO ACTIVITIES
2.1
Activity 1
a)
Find  cos 7 x dx
du
du
 7 and
 dx
dx
7
1
Hence  cos 7 x dx   cos u . du
7
1
 sin u  C
7
1
 sin 7 x  C
7
Let u  7 x, then
b)
Find  tan  5 x  7  dx
Let u  5 x  7, then
182
du
du
 5 and
 dx
dx
5
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
1
tan u . du
5
This compares with standard form 8.
1
 n sec u  C
5
1
 n sec  5 x  7   C
5
Hence  tan  5 x  7  dx 
c)
Find  x cot x 2 dx
Let u  x 2 , then
du
 x dx
2
  x cot x 2 dx  12  cot u du
This is an example of standard form 9.
 12 n sin x 2  C
d)
Find  sec  3 x  4  tan  3 x  4  dx
Let u  3 x  4, then du  3 dx and
du
 dx
3
  sec  3 x  4  tan  3x  4  dx  13  sec u tan u du
This is an example of standard form 14.
 13 sec u  C
 13 sec  3 x  4   C
e)
Evaluate  x 2 sec2 x3 dx
Let x3  u , then du  3x 2 dx and
du
 x 2 dx
3
1
sec2 u du
3
This is an example of standard form 12.
1
 tan u  C
3
1
 tan x3  C
3
Therefore  x 2 sec 2 x3 dx 
MAT1581
Mathematics I (Engineering)
183
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
f)
Find  x cosec2 x 2 dx
Let u  x 2 , then du  2 x dx and
du
 x dx
2
1
cosec2 u du
2
This is an example of standard form 13.
1
  cot u  C
2
1
  cot x 2  C
2
Therefore  x cosec 2 x 2 dx 
2.2
Activity 2
a)
Find  a sin x cos x dx
Let u  sin x, then du  cos x dx
Hence  a sin x cos x dx   a u du
b)

au
C
n a

a sin x
C
n a
Find  e3 cos 2 x sin 2 x dx
Let u  3cos 2 x, then du  6sin 2 x dx and 
Hence  e3 cos 2 x sin 2 x dx  
du
 sin 2 xdx
6
1 u
e du
6
1 3 cos 2 x
e
C
6
e3 cos 2 x

C
6

c)
Find  e x cos e x dx
Let u  e x , then du  e x dx
Hence  e x cos e x dx   cos u du
 sin u  C
 sin e x  C
184
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
d)
Find  tan 2 x dx
This does not seem to fit in one of the standard forms.
But, if we remember that tan 2 x  sec 2 x  1,
we see that
 tan x dx
2
   sec 2 x  1 dx
  sec 2 x dx   1 dx
Now we have standard form 12 in the first integral
= tan x  x  C
e)
 1 dx  sec2 x dx


 cos 2 x
 tan x  C
f)
Find  cos 2 x sin x dx
Let u  cos x, then du   sin x dx and  du  sin xdx
Therefore  cos 2 x sin x dx    u 2 du
u3
C
3
cos3 x

C
3

g)
Find  sec 4 3x tan 3x dx
This is difficult to recognise.
Remember that the derivative of sec x  sec x tan x
Rewrite the integrand  sec4 3 x tan 3 x dx   sec3 3 x  sec 3 x tan 3x  dx
Let u  sec 3 x, then du  3sec 3 x tan 3 x dx and
du
 sec 3 x tan 3 xdx
3
Therefore  sec 4 3x tan 3x dx   sec3 3x  sec 3x tan 3 x  dx
1 3
u du
3
u4

C
12
sec 4 3 x

C
12

MAT1581
Mathematics I (Engineering)
185
Module 7 Learning unit 6
INTEGRATION: Trigonometric functions
h)
Find  sin  2x  dx
Let u   2x  , then du  12 dx and 2 du  dx
Therefore  sin  2x  dx  2 sin u du
 2cos u  C
 2cos  2x   C
i)
Find  tan 3x sec2 3x dx
If f  x   tan 3x, then f '  x   3sec2 3x
Therefore  tan 3x sec2 3x dx


1
tan 3x 3sec2 3x dx

3
This is an example of standard form 8
1
 n sec3x  C
3

j)
 1  cos x dx   1 dx  cos x dx



 sin x
 sin x
 sin x
  cosec x dx   cot xdx
 n cosec x  cot x  n sin x  C
We can simplify our answer using logarithm rules
=n sin x  cosec x  cot 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.
186
MAT1581
Mathematics I (Engineering)
MODULE 7
LEARNING UNIT 7
INTEGRATION
The definite integral
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
CONTENTS
PAGE
1.
2.
3.
4.
DEFINITION ...................................................................................................... 188
EVALUATION OF A DEFINITE INTEGRAL ................................................. 188
RESPONSE TO ACTIVITY ............................................................................... 190
EXERCISE 1 ....................................................................................................... 191
MAT1581
Mathematics I (Engineering)
187
Module 7 Learning unit 7
INTEGRATION: The definite integral
1.
DEFINITION
You are already familiar with the procedure to evaluate an indefinite integral:
If g '  x   f  x  then  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:
a f  x  dx
 g b  g  a 
b
 g  x a
b
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:
 f  x  dx   g  b   C   g  a   C
b
a
 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  f  x  dx   f  x  dx .
b
a
a
b
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.
 f  x   a or f  x  a
Substitute the upper limit b for x in f(x), that is find f(b).
Substitute the lower limit a for x in f(x), that is find f(a).
Subtract f(a) from f(b).
b
(ii)
(iii)
(iv)
b
Example 1
a)
Calculate   4 x  2 dx
3
1
Since   4 x  2  dx  2 x 2  2 x  C
it follows from the definition that
  4 x  2  dx  2 x  2 x 
3
2
3
1
1
2
  2  3  2  3    2 1  2 1 

 

 12   0
2
 12
188
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 7
INTEGRATION: The definite integral
b)
Determine   4 x  2  dx
1
3
Since   4 x  2  dx  2 x 2  2 x  C
it follows from the definition that
  4 x  2  dx  2 x  2 x 
1
1
2
3
3
  2 1  2 1    2  3  2  3 

 

  0  12
2
2
 12
Note: Examples a) and b) illustrate that  f  x  dx   f  x  dx . Furthermore, the
b
a
a
b
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  cos 2x dx
2
0
Since  cos 2 x dx  12 sin 2 x  C
it follows from the definition that


2
0
d)
cos 2 x dx
  12 sin 2 x 0 2

  12 sin 2   2     12 sin 2(0) 
Evaluate 

0
Since 

0
2
2
  12 sin  
  12 sin 0
0
0
0
 sin x  3cos3x  dx
 sin x  3cos 3x  dx   cos x  sin 3x  C
it follows from the definition that

  sin x  3cos 3x  dx    cos x  sin 3x

2
2
0
0
   cos   2   sin 3   2      cos  0   sin 3  0  
    0 
 1
1
2
MAT1581
Mathematics I (Engineering)
 (1) 
   1
  1
  0  
1
189
Module 7 Learning unit 7
INTEGRATION: The definite integral
ACTIVITY 1
Calculate
  x  3x  dx
b)  x  x  1 dx
2
a)
2
1
0
2
2
1

 2 sin 2 x

dx
2

 0 1  cos x
c)
1
 4e dx
d)
3.
2
2x
-1
RESPONSE TO ACTIVITY
2
a)
 x3 3x 2 
x

3
x
dx
=


3  2 
1

1
3
  2  3  2 2   13 3 1 2 





2   3
2 
 3
 

8
 1 3
   6    
3
 3 2
2
2
16  36  2  9
6
41

6
5
6
6

b)
 x  x  1 dx  2   2  x  x  1 dx
0
1
2
2
1
0
2
2
1
3
 2

1   x  1 


2
3

1
3 0
1
  x 2  1 
1
6
3
3
1
2
2
   0   1  1  1 
6

1
 1  8
6
7

6
0

190
 

MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 7
INTEGRATION: The definite integral


c)
 2 2sin x .cos x
 2 sin 2 x


dx
dx


2


1  cos 2 x
 0 1  cos x
0
  n 1  cos 2 x 

2
0
  n 1  0   n 1  1 
  n 1  n 2
2
1
 n 2
 n
 0, 693
d)

1
2
-1
4e 2 x dx 
4 1 12

 2  e2 x dx
1 2 1
 2 e 2 x 
1
2
1
 2 e  e 2 
1
1

 2 e  2 
 e 
 5.166
4.
EXERCISE 1
Evaluate each of the given definite integrals:
a)
c)
  x  3 dx
 4 x  x  3 dx
2
2
b)
1
2
3
2
d)
1
1
e)
  1
1

 3  dx
  x2
x 
3 
g)



6
10
f)
  x  x  2  dx
 x 1  x  dx
1
2
1
0
1

3
h)
e  2 dx
x
2
3
dx
x2
2 2
3
 sin x dx

4
2
ANSWERS
a)
f)
1
3
4.99
5
b)
g)
MAT1581
Mathematics I (Engineering)
2
3
0.693
4
c)
7.5
h)
0.707
d)
0.3
e)
1
1
9
191
Module 7 Learning unit 7
INTEGRATION: The definite integral
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.
192
MAT1581
Mathematics I (Engineering)
MODULE 7
LEARNING UNIT 8
INTEGRATION
Areas
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
CONTENTS
PAGE
1.
2.
3.
4.
4.1
4.2
5.
INTRODUCTION ............................................................................................... 194
SUMMATION AND THE DEFINITE INTEGRAL .......................................... 195
THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE ................. 196
RESPONSES TO ACTIVITIES.......................................................................... 205
Activity 1 ............................................................................................................. 205
Activity 2 ............................................................................................................. 206
EXERCISE 1 ....................................................................................................... 208
MAT1581
Mathematics I (Engineering)
193
Module 7 Learning unit 8
INTEGRATION: Areas
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:
Velocity
Time
To find the distance travelled, the area under the graph can be calculated by dividing the
area into geometric shapes:
Velocity
Time
The area of rectangles and triangles can be found easily.
Now consider:
Velocity
v(t)
Time
194
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
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.)
Y
f( x)
0
a
b
X
Figure 1
Divide the interval into n equal parts each of length x.
Y
x
0
a
b
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.
MAT1581
Mathematics I (Engineering)
195
Module 7 Learning unit 8
INTEGRATION: Areas
In figure 3 a representative strip and its approximating rectangle are shown.
Y
Pi (xi;yi)
R
0
a
S
M x N
x
xi
b
X
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 xaxis 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 i th strip = height of strip  length of strip
 yi  x
 yi 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
n
y1 x  y2 x  y3 x  .....  yn x   yi x
i 1
[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
196
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
n
Area  lim  yi x
n 

b
a
i 1
y dx
  f  x  dx
b
a
Thus the summation process is a special integration process. In fact, the elongated ssymbol  is meant to indicate a type of summation.
3.
THE DEFINITE INTEGRAL AS THE AREA UNDER A CURVE
Geometrically,  f  x  dx is interpreted as the area between the curve f(x), the x-axis
b
a
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 f  x   0, then  f  x  dx  0 :
b
a
Y
f( x)
0
a
b
X
If f  x   0, then  f  x  dx  0 and area =  f  x  dx :
0
b
b
a
a
a
b
X
f( x)
Y
MAT1581
Mathematics I (Engineering)
197
Module 7 Learning unit 8
INTEGRATION: Areas
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 xand y-axes.
We draw a sketch of the required area:
3
We can find the equation of the straight line AB using ordinary geometry as y   x  3
4
Method 1:
Using the formula
Area of AOB  12  width  height 
43
2
 6 units 2

198
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
Method 2
Using vertical strips:
Sketch the required area and a representative strip.
Area of representative strip = height  width
 y  x
 3

   x  3  x
 4

Therefore the required area A can be calculated with the definite integral:
A
x4
x 0
y dx
4
  3

    x  3  dx

0  4
4
 3x 2

 
 3x 
 8
0
 3  4 2

 
 3  4     0  0
8


  6  12 
 6 units 2
MAT1581
Mathematics I (Engineering)
199
Module 7 Learning unit 8
INTEGRATION: Areas
Method 3:
Using horizontal strips
Sketch the required area and a representative strip.
Area of representative strip A = x . y
Required area A  
y 3
y 0
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.
3 x
y
3
4
3x
  y  3
4
4y
x  
4
3
Therefore the required area is
A
y 3
y 0
x dy
3
  4y

   
 4  dy
3

0 
3
 4 y2

 
 4 y
 3 2
0
3
 2 y2

 
 4 y
 3
0
3
 2  3 2

 
 4  3 
3

 0
 6  12
 6 units 2
200
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
Example 2: Calculating distance from a velocity time graph
The velocity v of a body t seconds after a certain instant is ( 2t 2  5 ) m/s. Find by
integration how far it moves in the interval from t = 0 to t = 4 s.
Since 2t 2  5 is a quadratic expression, the curve v = 2t 2  5 is a parabola cutting the vaxis at v = 5. The distance travelled is given by the area under the v/t curve.
4
4
0
0


Thus distance =  v dt   2t 2  5 dt
4
 2t 3


 5t 
 3
0
3

 2(4)3
  2 0

 5(4)   
 5  0 

 3
  3
128


 20   0
 3

 62.67 m
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.
MAT1581
Mathematics I (Engineering)
201
Module 7 Learning unit 8
INTEGRATION: Areas
Example 3: Area between a curve and the axes
Find the area bounded by the curve y  x 2 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.
y = x2
The required area is shown (shaded) above.
Area of representative strip A = x .y
 Required area A  
x 2
x 1
y dx
2
  x 2 dx
1
2
 x3 
 
 3 1
 2   1
3

3
3
3
8 1
 
3 3
7

3
1
 2 square units
3
202
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
ACTIVITY 1
Find the area bounded by the curve y  sin x and the x-axis between x  0
and x   .
Remember to check the response on page 205.
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.
Area of representative strip = y dx
Area  
x 
x 0
y dx  
x  2
x 

2
0

  sin x dx  
y . dx
sin x dx
   cos x 0    cos x 

2
because  cos x   cos x
   cos   cos 0   cos 2  cos  
    1  1  1   1 
 22
 4 square units
MAT1581
Mathematics I (Engineering)
203
Module 7 Learning unit 8
INTEGRATION: Areas
+The necessity for a sketch and the above method of evaluation is immediately evident if
we evaluate 
x  2
x 0

x  2
x 0
y dx  
2
0
y dx  
2
0
sin x dx .
sin x dx
   cos x 0
2
   cos 2  cos 0
  1  1
0
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 f  x   x 2  2 x  3 and the x-axis
between x  2 and x = 4.
Remember to check the response on page 206.
Example 5: Area between two curves
Calculate the area enclosed by the parabolas y  6 x  x 2 and y  x 2  2 x .
Sketch the parabolas on the same set of axes.
Find the intersection of the parabolas by setting the equations equal:
x2  2x  6x  x2
2 x2  8x  0
x2  4x  0
x  x  4  0
x  0 or x  4
Find the y-values by substituting into one of the equations.
204
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
Thus the intersections of the parabolas are at (0;0) and (4;8).

 
Length of a typical strip: y1  y2  6 xk  xk 2  xk 2  2 xk

Required area    y1  y2  dx
4
0
4


  8 x  2 x 2 dx
0
  4 x 2  23 x3 
 4(16)  23
4
0
 64 
64
3
 21.33 units 2

4.
RESPONSES TO ACTIVITIES
4.1
Activity 1
Sketch the required area.
Area of representative strip : A  y.x
Required area A  
x 
x 0
y dx

  sin x dx
0
   cos x 0

   cos   cos 0
   1  1
 2 square units
MAT1581
Mathematics I (Engineering)
205
Module 7 Learning unit 8
INTEGRATION: Areas
4.2
Activity 2
1. We first sketch the area.

A1 : A1   cos x dx
2
0
  sin x 0 2

 1 square unit

A2 : A2   cos x dx
2
  sin x 

2
 1square unit
 Total area  A1  A2
 11
 2 square units
As before:

 cos x dx  0 which is the algebraic sum of the two areas.
0
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.
206
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.
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
x-intercepts  Put f  x   0 :
x2  2 x  3  0
 x  3 x  1  0
 x  1 or 3
y -intercept  Put x  0 :
y  3
Turning point  Use differentiation or the formulae from school
f ' x  2x  2
and f 1  12  2 1  3
 2x  2  0
x 1
 4
 Coordinates of turning point 1; 4  .
Area 1  
x 1
x 2

1
2
y dx
 x  2 x  3 dx
2
1
 x3

   x 2  3x 
3
 2
  31  1  3   38  4  6  
 2   2  
  1      
 3   3  
1
 2 units 2
3
MAT1581
Mathematics I (Engineering)
207
Module 7 Learning unit 8
INTEGRATION: Areas
Area 2  
x 3
x 1
y dx
   x 2  2 x  3 dx
3
1
3
 x3

   x 2  3x 
3
 1
  273  9  9    31  1  3 

 2 
   9    1  
 3 

2
 10
3
2
 10
3
Area 3  
x4
x 3
y dx
   x 2  2 x  3 dx
4
3
4
 x3

   x 2  3x 
3
3
  643  16  12    273  9  9  
 2

  6   9  
 3

 1
 2 
 3
Area  Area 1  Area 2  Area 3
1
 15 square units
3
5.
EXERCISE 1
a)
Find the area enclosed between the x-axis and y  4  x 2 .
b)
Find the area between the x-axis and y  x 3 from x = 1 to x = 1
c)
Find the area bounded by the parabola x  4  y 2 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 y  3  2 x  x 2 and the x-axis.
e)
Calculate the area between the curve y  x3  x 2  3x  1 , the x-axis and the lines
3
x = 1 and x = 3.
208
MAT1581
Mathematics I (Engineering)
Module 7 Learning unit 8
INTEGRATION: Areas
ANSWERS
a)
d)
10.67 square units
1
units 2
2
2
10 units 2
3
10, 67 units 2
e)
12 units 2
b)
c)
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.
MAT1581
Mathematics I (Engineering)
209
POST-TEST: INTEGRATION
1.
Determine


x
 a     x   
2


2 
 dx
 x  
(b)
  x  1 x  1 dx
c
 3x  4 x  7
dx

x

2
Determine z if z    9 y 3  11 y 2  y  3 dy , given that when y  1, z  2.
2.
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
ds
displacement, that is v 
dt
(a)
Derive an expression for s.
(b)
Given that when t  0 the displacement s  0 , find the value of the
constant of integration.
4.
An object moves with variable acceleration a given by the formula a  3t 2 .
(a)
Find an expression for the velocity of the object.
(b)
Find an expression for the displacement of the object.
5.
Find
a 





c




(e)





(g)
  e  1 e dx
6.
x 2 dx
1 x
dx
3
2  3x
2
2x2  3
2 x3
 3x  7
3
x
2
b
 x x  4 dx
d
 x 19  17 x dx
(f)
 3t.
2
2

t2  3
 dt
3
x
Determine the following integrals, using partial fractions where necessary to
simplify the integrand:
 x2  2 x
dx
(a) 
2
x

1



 x  5x  4
dx
(c) 
x2

3
(b)
2
MAT1581
Mathematics I (Engineering)
(d)



x 1
x  2x  4
2
dx
 9 x2
dx
 3
3
 ( x  2)
210
Module 7
INTEGRATION: Post-test
 x2
dx
4 3
 x 2

2 x 2  9 x  35
dx
(g) 
  x  1 x  2  x  3
(e)
7.
(f)
 11  3x dx
 2
 x  2x  3
Evaluate
 1  x 
dx

x

2
(a)  x  2 x dx
2
8.
4
(b)
Determine
(a)  tan 2x dx
(b)
(c)  5 esin x cos x dx
(e)
9.
3 sin 2 3 x .cos 3 x dx
(d)
 cos ec x
dx

 sin x
e
(f)
tan 2 x
sec 2 x dx
 sec3x tan 3x dx

 5  2sec3x
Evaluate the following integrals:
(a) 
3
1
2
 x  2 x dx
2
3
(b)  x x 2  1 dx
1

(c)  ecos x sin x dx
2
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.
MAT1581
Mathematics I (Engineering)
211
INTEGRATION: POST-TEST SOLUTIONS
1.
a 






2 
 dx
 x  
 1 x
1 
   x 2   2 x 2  dx
2


 x 
x

2
3
1
1 x2
x2
 3  .  2. 1  C
2 2
2
2
x2
2 x3 x 2
 4 x C
3
4
 x  1 x  1 dx

(b)



Simplify integrand before integration


=  x  2 x  1 dx
3
x2 2x 2

 3  xC
2
2
3
x2 4x 2


 xC
2
3
As the question was written in surd (root) form, we write the answer as

c
x 2 4 x3

 xC
2
3
 3x 2  4 x  7
dx

x

  3x 2 4 x
7 
 


 dx
x
x
 x
1 1
2
  2 1
1 
   3 x 2  4 x  7 x 2  dx


1
2
 3
1 
   3 x 2  4 x  7 x 2  dx




5
3
1
3x 2
4x 2
7x2
5
2

5
2
3
2
3
2

C
1
2
1
2
6x
8x
14 x


C
5
3
1
6 x5 8 x3

 14 x  C
5
3
z    9 y 3  11 y 2  y  3 dy

2.
MAT1581
Mathematics I (Engineering)
212
Module 7
INTEGRATION: Post-test solutions
9 y 4 11y 3 y 2


 3y  C
4
3
2
given y  1, z  2
9 11 1
2    3C
4 3 2
9 11 1
C  2   3
4 3 2
24  27  44  6  36

12
5

12
Thus the answer is

z
9 y 4 11y 3 y 2
5


 3y 
4
3
2
12
3.
(a)
Given that v  u  at , and that v 
ds
, we can write
dt
ds
 u  at
dt
so that
s    u  at  dt
 s  ut  12 at 2  C
where C is a constant of integration.
(b)
We are given that when t  0, s  0, and substituting these values gives
s  ut  12 at 2  C
0  u  0   12 a  0   C
2
0C
Therefore s  ut  12 at 2
4.
Given that a  3t 2 .
(a)
v   a dt   3t 2 dt  t 3  C
(b)


s   v dt   t 3  C dt  14 t 4  Ct  D
where C and D are constants of integration
MAT1581
Mathematics I (Engineering)
213
Module 7INTEGRATION: Post-test solutions
5.
a 





x 2 dx
1  x3

1  3 x 2
dx
3  1  x 3
1 1  x 
 .
C
3
 b
2
1
2
3

1
1
2
2
1  x3   C

3
1
 x x  4 dx  2  2 x x  4 dx
2
2
1  x  4

C
3
2
2
3
2
2
 x  4  C

3
2
2
3
c




dx
2  3x 2
Let u  2  3x 2 then du  6 x.
This integral cannot be solved with the methods we discussed thus far;
You will learn suitable methods in Mathematics II.
(d)
 x 19  17 x dx
2
Let u  19  17 x 2
then, du  34 x dx
and  x 19  17 x dx  
2
1
34
1
2
 u du
3
  341  32  u 2  C

3
2
19  17 x   C
2
51
19  17 x 
2 3


(e) 



214
51
C
2x2  3
2 x3
 3x  7
3
2 x3
 3 x  7 then du  (2 x 2  3) dx .
Let u 
3
MAT1581
Mathematics I (Engineering)
Module 7
INTEGRATION: Post-test solutions

and 



2x2  3
2 x3
 3x  7
3
 u

1
2
du
1
2
 2u  C
1
 2 x3
2
 2
 3x  7   C
 3

2 x3
 3x  7  C
3
2
(f)
 3t.  t  3 
2
3
dt  32
 2t.  t
2
3
 3 2 dt

1
3 2 2
t 3 2 C
 .
2 1
3

C
t2  3

(g)

  e  1 e dx
x
2
x
Let u  e x  1, then du  e x dx


2
Therefore  e x  1 e x dx   u 2 du

u3
C
3
 e  1  C

3
x
3
6.
(a)
2
 x2  2 x
 x  2x

dx
dx


2
 x2  2 x  1
  x  1
We can use partial fractions or manipulate the integrand as follows:
2
 x  2 x  1  1

dx
2
 x  2x  1
2
 ( x  2 x  1)  1

dx
 x2  2 x  1
2
1
 ( x  2 x  1)
dx 
dx
 2
 2
 x  2x  1
 x  2x  1

  1 dx

MAT1581
Mathematics I (Engineering)

1
dx

2
  x  1
215
Module 7INTEGRATION: Post-test solutions
   x  1 dx
2
x
x

1
1

 x  1
x
(b)
x 1
 x  11



x2  2 x  4
 12
 (2 x  2)  x  2 x  4 
C
C
dx
2



1
2 dx
1
1 2
 . x2  2x  4 2  C
2 1
 x  5x  4
dx

x2

  x  5  4 x 2 dx
3
(c)
2


x2
4

 5x   C
2
x
(d)
 9 x2
dx
 3
3
 ( x  2)

 3 3 x 2 x 3  2
 dx
3
 x  2  C
 3.
2
3

(e)
2
3

2 x 2
3

2
C
 x2
dx
4 3
 x 2



1
1
3 x 2 x3  2 4 dx

3




3
1 4
 . x3  2 4
3 3
3
4
 4 x3  2  C
9
216
MAT1581
Mathematics I (Engineering)
Module 7
INTEGRATION: Post-test solutions
(f)
 11  3x dx
 2
 x  2x  3
Resolve the integrand in partial fractions
A  x  3  B  x  1
11  3 x
11  3 x
A
B




x  2 x  3  x  1 x  3 x  1 x  3
 x  1 x  3
2
Equating numerators gives
11  3 x  A  x  3  B  x  1
Let x  1: 8  4 A
A 2
Let x  3 : 20  4 B
B  5
11  3 x
dx
Hence 
 2
 x  2x  3
2
5

dx  
dx


 x3
 x 1
 2n x  1  5n x  3  C
 x  12
 n
C
5
 x  3
(g)
by the laws of logarithms

2 x 2  9 x  35
 x  1 x  2 x  3 dx




Resolve the integrand in partial fractions
2 x 2  9 x  35
 x  1 x  2  x  3

A
B
C


x 1 x  2 x  3

A  x  2  x  3  B  x  1 x  3  C  x  1 x  2 
 x  1 x  3 x  2 
Equating numerators gives
2 x 2  9 x  35  A  x  2  x  3  B  x  1 x  3  C  x  1 x  2 
Let x  1:  24  6 A
A 4
Let x  2 :  45  15 B
B  3
Let x  3 :10  10C
C 1
MAT1581
Mathematics I (Engineering)
217
Module 7INTEGRATION: Post-test solutions

2 x 2  9 x  35
Hence 
dx
  x  1 x  2  x  3
4
3
1

dx  
dx  
dx



 x 1
 x2
 x3
 4 n x  1  3 n x  2  n x  3  C
 x  1  x  3  C
 n
3
 x  2
4
by the laws of logarithms
7.
(a)
 x  2 x dx
  x 2 1  2 x 2  dx
2
4
1  2 x  dx
  x2

2
1
2 2
x 1  2x
dx


1

 4 x  1  2 x 2 2 dx
4 


1
1 2
 . 1  2x2 2  C
4 3
3
1
1  2 x2  C

6




 1  x 
dx

x

2
(b)
3
 1  2 x  x2

dx
1


x2
1
3
  1

   x 2  2 x 2  x 2  dx




1
3
5
4 2
2 2
2
 2x  3 x  5 x  C
8.
(a)
 tan 2 x dx  12  2 tan 2 x dx  12 n sec 2 x  C
(b)
If f  x   sin x, then f '  x   cos x
Therefore  5 esin x cos x dx  5 esin x  C
(c) 3 sin 2 3 x cos 3 x dx 
(d)
218
e
tan x
sin 3 3 x
C
3
sec 2 x dx  e tan x  C
MAT1581
Mathematics I (Engineering)
Module 7
INTEGRATION: Post-test solutions
 cos ec x
dx   cosec2 x dx   cot x  C
(e) 
 sin x
(f)
9.
(a)
 sec3x tan 3x dx

 5  2sec3x
Let u  5  2sec 3x, then du  6sec 3x tan 3x dx
sec 3x tan 3x
1 1
Therefore 
dx  

 du
 5  2sec 3x
6 u
1
 n u  C
6
n 5  2sec 3x

C
6
Let I    x 2  2  x dx
3
3
1
du
 2x
dx
The values of the limits must also be changed to correspond with the
new variable u :
Let u  x 2  2, then
Therefore if x  1 then u  12  2  1
and if x =3 then u  32  2  7
1 7
 I   u 3 du
2 1
7
1 
  u4 
 8  1
1
  2401  1
8
 300
(b)
2
Put I   x x2  1 dx
1
MAT1581
Mathematics I (Engineering)
Let u  x 2  1, then
du
 2x .
dx
219
Module 7INTEGRATION: Post-test solutions
Therefore if x  1 then u   1  1  0
2
and if x =2 then u   2   1  3
2
1 2 2
x  1.2 x. dx
2 1
1 3 1
  u 2 du
2 0
I 
3
1 u 2 
  3 
2  2 0
3

1 32
3
3 0 2
3
1 1 12
 .3 .3
3
 3

(c)


Put I   ecos x sin x dx
2
u  ecos x
du
Then
  sin x ecos x
dx
and du   sin x ecos x dx
Let
Now substitute in the original integral.
b
I    du
a
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.
b
I    du
a
  u a
b

  ecos x  
2
 cos 2 
   e    e



1
0
 e  e
1

1
e
cos 
10.
8  2 y  y2  x
y -intercepts:  4  y  2  y   0
y  4 or y  2
220
MAT1581
Mathematics I (Engineering)
Module 7
INTEGRATION: Post-test solutions
Area  
3
1
8  2 y  y  dy
2
3
y3
 8y  y 
3 1
2
  24  9  9    8  1  13 
 24   6 23 
2
3
3
x  6 x2  8x  x x2  6 x  8
 30
11.


 x  x  4  x  2 
MAT1581
Mathematics I (Engineering)
221
Module 7INTEGRATION: Post-test solutions
2


Area   x3  6 x 2  8 x dx  
0
4
2
 x  6 x  8 x  dx
3
2
2
 x4

x4

 2 x3  4 x 2    2 x3  4 x 2 
4
 4

0
4
2
 4  4
8
12.
2x2  x  0
x  2 x  1  0
x  0 of/or x 
Area  
0
3
1
2
 2 x  x  dx    2 x  x  dx    2 x  x  dx
0
2
1
0
2
2
1
2
0
2
2
2
 2 x3 x 2 
 2 x3 x 2 
 2 x3 x 2 

  
  
 
2 
3
2 1  3
2 1
 3

3
2
2
1   1 1   16
1 1

   18  4          2   
2   12 8   3
2 8

1 1 81
 22 

2 24 24
 25
222
11
12
MAT1581
Mathematics I (Engineering)
Module 7
INTEGRATION: Post-test solutions
y  4 x 2  3x  1
13.
4 x 2  3x  1  0
 4 x  1 x  1  0
x
2

1
of/or x  1
4

Area   4 x 2  3x  1 dx
1
2
 4 x3 3x 2



 x
2
 3
1
 32
 4 3 
   6  2      1
 2
 3 2 
2  1
 2   1 
3  6
3
5
6
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.
MAT1581
Mathematics I (Engineering)
223