Download Note Chapter 4 (Integral Calculus)

BQT 133-Business Mathematics
Teaching Module
CHAPTER 4 : INTEGRAL CALCULUS
4.1
INTRODUCTION
Integration is the reverse process of differentiation. For that reason, it is also known as antiderivative which means that if given derivative of a function we can work backwards to find the
function from which it is derived.
For example: If
d
F ( x)  f ( x) , then F(x) is the anti-derivative of f(x) and the process of finding
dx
F(x) is called integration. It is written as
 f ( x)dx  F ( x)  c , where the symbol  f ( x)dx is the
integral of f(x), f(x) is called the integrand and c is the constant of integration.
Definition 4.1 : Anti Derivatives
Given a function f(x) an antiderivative of f(x) is any function F(x) such that
F ( x)  f ( x)
Beside F(x), c is also part of the product of the integration because of the explanation given
below:
d 3
( x  1)  3 x 2
(i)
dx
d 3
( x  1)  3 x 2
(ii)
dx
From (i) and (ii) we can conclude that the functions F ( x)  x 3  1 and F ( x)  x 3  1 are both
anti-derivatives of f ( x)  3x 2 . In that case, we can also conclude that the function F ( x)  x 3
plus any constant (positive or negative), can also be the anti-derivatives of the function
f ( x)  3x 2 . Therefore, if we replace this constant with the letter c, then, we can say that
F ( x)  x 3  c is the anti-derivative of f ( x)  3x 2 . Hence, the anti-derivative is written as
 3x dx  x
2
3
 c and this integral is known as indefinite integral because of the indefinite value
of c. The value of c could be determined if the corresponding values of x and y are given.
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Definition 4.2 : Indefinite Integral
If F(x) is any anti-derivative of f(x) then the most general anti-derivative of f(x) is called an
indefinite integral and denoted,
 f ( x)dx  F ( x)  c , c is any constant
In this definition

is called the integral symbol, f(x) is called the integrand, x is
called the integration variable and the “c” is called the constant of integration.
Table 4.1 which shows the relationship between derivative and indefinite integral formulas for
some elementary functions.
Derivative Formula
d 2
x  2x
dx
d x
e  ex
dx
Equivalent Integral Formula
 
 2xdx  x  c
 e dx  e  c
2
 
x
x
Table 4.1: Derivative and Integral Formulas of Elementary Functions
Like differentiation, integration has many practical applications. The obvious are, for examples,
in kinematics, estimating populations growth, measuring the arc length, surface area,
volume of solid and center of gravity.
4.2
STANDARD INTEGRAL
f (x)
xn
1
a
ex
ax
1
x
 f ( x)  dx
x n 1
 C , n  1
n 1
xC
ax  C
e x +C
ax
C
ln a
ln x  C
From the above table , we can use this basic function to solve the integral of polynomial function,
integral of exponential functions, integral of logarithmic functions and integral of trigonometric
functions.
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4.3
BQT 133
INDEFINITE INTEGRALS
In this section we shall prove two basic properties of indefinite integrals. The basic properties of
indefinite integrals are stated in the following theorem.
Theorem 4.1: Basic Properties of Indefinite Integrals
1.
The constant factor k can be taken out from an integral, that is
 kdx  k  f ( x)dx ,
2.
where k is a constant
The integral of a sum or differences equals the sum or difference of
the integrals, that is
  f ( x)  g ( x)dx   f ( x)dx   g ( x)dx
4.3.1 Integral of Polynomial Functions
Properties of the Integral of Polynomial Functions
x n 1
c
n 1
1.
n
 x dx 
2.
 kdx  kx  c ,
(General form)
where k is any number
Example 4.1
Determine the following integrals:
(a)
x
7
(b)
dx
 2x
8
dx
(c)
1
x
6
dx
(d)

3x dx
Solutions:
(a)
7
 x dx 
x 7 1
x8
c 
c
7 1
8
2 x 81
 2 x 7
(b) 2 x dx  2 x dx 
c 
c
 8 1
7
1
x 61
 x 5
(c)  6 dx   x 6 dx 
c 
c
 6 1
5
x

(d)

8

8
3x dx  3  x dx  3  x
12
1
1
2
3
2
3
x
x
2 3 2
 3
c  3
c 
x c
1
3
3
1
2
2
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Exercise 4.1:
(a)

4
(b)
x dx

1
6
x2
(c)
dx
Example 4.2
Determine the following integrals:
(a)  (2 x  5 x 3  4 x 2 )dx
1
 x 3 2 dx

2
(d)   3x 3 dx
(b)  (3x  5) 2 dx
 2x  1 
(d)   4 dx
 x 
2
2

(c)   3x   dx
x

Solutions:
(a)  (2 x  5 x 3  4 x 2 )dx



 2 xdx  5 x dx  4 x dx
 2 xdx  5 x 3 dx  4 x 2 dx
3
2
 x2
 2
 2
  x4   x3 
  5   4   c
  4   3 
5
4
 x2  x4  x3  c
4
3
(b)  (3x  5) 2 dx
  (9 x 2  30 x  25)dx
  9 x 2 dx   30 xdx   25dx
 9 x 2 dx  30  xdx  25 1dx
 x3 
 x2 
 9   30   25( x)  c
 3 
 2 
 3x 3  15 x 2  25 x  c
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(c)

BQT 133
2
2

 3x   dx
x

4 

   9 x 2  12  2 dx
x 

  9 x 2 dx   12dx  
4
dx
x2
 9  x 2 dx  12  1dx  4  x  2 dx
 x3 
 x 1 

 9   12 x  4
 3 
 1 
4
 3 x 3  12 x 
x
(d)
 2x  1 
dx
x4 
2x
1
  4 dx   4 dx
x
x
 
 2 x 3 dx   x  4 dx
 x  2  x 3
 
 2

2

 3
1
1
 2  3
x
3x
Exercise 5.2:
(4 x  3) 2
(a) 
dx
x
(b)

(3x  2) 3
3
x2
dx
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4.3.2 Integral of Exponential Functions
Formula of the Integral of Exponential Functions
1.
e
2.

3.
e
x
dx  e x  c
e axb dx 
u
(Integral of Natural Exponential Function)
e axb
 c ; where a any number
a
du  e u  c
 g ( x)(e
or
g ( x)
)dx  e g ( x )  c
(by using Substitution Rule)
Example 5.3
Solve the following integrals:
(a)  e 2 x dx
(b)  e 2 x 3 dx
(d)
 e
2x

2
 e  2 x dx
(e)
 e
2x

(c)
 e
(f)
e
2
 3 dx
5x
x

 e 3 x dx
 e 3 x dx
Solutions:
(a)  e 2 x dx 
(b)
e
(c)
 e
(d)
 e
2 x 3
5x
2x
e2x
c
2
e 2 x 3
dx 
c
2
e
3 x
dx   e

5x
dx   e
3 x
e 5 x  e 3 x 
e 5 x e 3 x
c 
dx 


c
5   3 
5
3
2
 e  2 x dx


  e 4 x  2  e 4 x dx
  e 4 x dx   2dx   e  4 x dx

 e 4 x
e4x
 2 x  
4
 4

  c


e4x
e 4 x
 2x 
c
4
4
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(e)
 e
2x

2
BQT 133
 3 dx 

 e
e
4x
4x
4x

 6e 2 x  9 dx
dx 
 6e
2x

dx  9dx
2x
e
6e

 9x  c
4
2
e4x

 3e 2 x  9 x  c
4

(f)

e x  e 3 x dx   e 4 x dx 
Exercise 5.3
e5x
(a)  2 x dx
e
e4x
c
4
(b)

e 4 x  e8x
dx
e2x
(c)

e 2x  e5x
dx
e9x
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4.3.3 Integral of Logarithmic Functions
Formula of Integral of Logarithmic Functions
1.
2.
3.
4.
 xdx  ln x  c
a
 ax  bdx  ln ax  b  c
1
1
 u du  ln u  c
g ( x)
 g ( x) dx  ln( g ( x))  c
or
(by using Substitution Rule)
ax
x
(
a
)
dx

c

ln a
Example 4.4
Find:
1
dx
(a) 
x3
(b)  3 x dx
(c)
7
 5  xdx
Solutions:
1
dx  ln x  3  c
(a) 
x3
(b)  3 x dx 
(c)
3x
c
ln 3
7
1
1
 5  xdx  7 5  x dx  7 ln 5  x   1  c  7 ln 5  x  c
Exercise 5.4
3
dx
(a) 
2x  3
(b)

5
dx
x
(c)
3
4 x 2
dx
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BQT 133
Many of the indefinite integrals are found by “reversing” derivative formulae. Table 4.2 give the
basic integral formulae for some standard functions.
Corresponding Derivative Formula
d
Indefinite Integral
dx
 `1dx  x  C
d  x n 1 
n

  x , n  1
dx  n  1 

[ x]  1
x n dx =
x n1
 C , n  1
n 1
 x dx  ln x  C
 e dx  e  C
d
1
[ln x ] 
dx
x
d x
x
[e ]  e
dx
d kx
[e ]  kekx
dx
d x
[a ]  a x ln a
dx
1
x
x

e kx dx 
e kx
C
k

a x dx 
ax
C
ln a
Table 4.2: Basic Derivatives and Integral Formulae
4.4
DEFINITE INTEGRALS
We have learnt about the indefinite integral of f which is denoted by
 f ( x)dx . In this section we
will discuss about the definite integral of f for continuous function on closed interval a, b ,
denoted by

b
a
f ( x) dx . The definition of the definite integral is summarized in a theorem called
Fundamental Theorem of Calculus.
Definition 4.2 : Definite Integral
A definite integral is an integral

b
a
f ( x) dx
With upper and lower limits it mean f(x) continuous on closed interval [a,b] and F(x) is the antiderivative (indefinite integral)of f(x) on [a,b] then

b
a
f ( x) dx  F ( x)a  F (b)  F (a)
b
(Fundamental Theorems of Calculus)
From this definition, it is clear that the function f must be continuous on the interval a  x  b so
that it is differentiable on the interval. The number a is the lower limit and b is the upper limit of
the integration.
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BQT 133
Note that there is no constant in definite integral, therefore definite integral is always in number.
This is because the constant c is eliminated as shown below.
If
 f ( x)dx  F ( x)  c , where c is a constant, then
 f ( x) dx  F ( x)  F (b)  c  F (a)  c
b
b
a
a
 F (b)  F (a)
Theorem 4.2: Basic Properties of Definite Integrals
If f(x) and g(x) are continuous functions on the interval [a, b], then
(a)


a

b
a
b
(b)
a
(c)
a
f ( x) dx  0 , if f(a) exists
a
f ( x) dx    f ( x) dx
b
b
kf ( x) dx  k  f ( x) dx
a
b
(e)
 k dx  k (b  a)
 f ( x) dx   f ( x) dx  
(f)
a  f ( x)  g ( x) dx  a
(d)
a
b
c
b
a
a
c
b
b
f ( x) dx , where a  c  b
b
f ( x) dx   g ( x) dx
a
Example 4.7
Solve the following integrals:

1  3x 
25
(d)  dx
1 x
(a)
2
2
dx
x
(b)
 2x  1 
1  x 4 dx
3
(c)
1
2
0

e 4 x  e8 x
dx
e2x
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BQT 133
Solutions:
(a)

2
1
2
2

 3 x   dx
x

2
4 
   9 x 2  12  2 dx
1
x 

2
2
  9 x 2 dx   12dx  
1
1
2
1
2
2
4
dx
x2
2
 9  x 2 dx  12  1dx  4  x  2 dx
1
1
1
2
4

  3 x 3  12( x)    (3(8)  12(2)  2)  (3  12  4)  11
x1

(b)
3
 2x  1 
dx
x4 
3 2x
3 1
  4 dx   4 dx
1 x
1 x
 
1
3
3
 2  x 3 dx   x  4 dx
1
1
3
3
 x 2 
x 3
 
 2
  2 1 3 1
3
1 
1  98
 1
 1 1 
  2  3       1  
3  81
3 x  1  9 81  
 x
(c)
1
2
0

e 4 x  e8 x
dx
e2x

1
2
 e
0
2x

 e 6 x dx
1
 e2x e6x  2

 

2
6

0
 1 2 1  1 6 1    1 2 ( 0 ) 1 6 ( 0 ) 
  e 2  e 2    e
 e 
2
 2
6
6




(d)

2
1
1
1
1 1 1
1
2
e  e3    e  e3 
2
6
2 6 2
6
3
21
5
2
dx  5  5 ln x1  (5 ln 2)  (5 ln 1)  3.4657  0  3.4657
1 x
x
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BQT 133
4.5 TECHNIQUES OF INTEGRATION
4.5.1 Integration by Substitution
This method is the first to be considered, whenever we try to obtain any integral. The purpose of
this method is to change the integrand into an expression of basic integral forms. In principle,
the process of integration by substitution can be done through five steps as follows:
Step 1:
Make appropriate choice of u, let u  g (x)
Step 2:
Obtain
Step 3:
Substitute
u  g (x) ,
du
 g ' ( x)
dx
du  g ' ( x)dx
After this stage, the whole integral must be in terms of u. This means
no more term in x can remain. If this step fails, we need to make
another appropriate choice of u.
Step 4:
Evaluate the integral obtained in terms of u.
Step 5:
Substitute u with g(x), so that the final answer will be in terms of x.
Some integral forms that can be evaluated using integration by substitution, are
(i)
 f (ax  b)
(ii)

(iii)
 f ( x) f ' ( x)
dx
f ' ( x)
dx
f ( x)
dx
We shall illustrate each case above with examples.
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(i)
BQT 133
Function of a Linear Function of x
We are very required to integrate functions like those in standard list, but where x is replaced by a
linear function of x, e.g.
 (5x  4)
6
dx , which is very much
x
6
(5 x  4) . If we put u to stand for (5 x  4) , the integral becomes
dx except that x is replaced by
u
6
dx and therefore we can
complete the operation, we must change the variables, thus:
dx
du
u 6 dx  u 6
du


dx
du
dx 1
 5 , therefore

can be found from substitution u  (5 x  4) for
and the
du
dx
du 5
integral becomes:
1 6
1 u7
6
6 dx
61
u dx  u
du  u  du 
u du  
c
du
5
5 7
5
Now




Finally, we must express u in terms of the original variables, x, so that:
(5 x  4) 7
(5 x  4) 7
(5 x  4) 6 dx 
c 
c
57
35

Example 4.8
Evaluate the following integrals by using the substitution u  4x  1.
(a)  (4 x  1) dx
(b)  (4 x  1) 2 dx
Solutions:
du
1
 4  du  dx . Substituting u and du give us
dx
4
1
 (4 x  1) dx  4  u du
1
 u2  c
8
1
 (4 x  1) 2  c
8
(a) If u  4x  1, then
Remarks
By integrating directly will get
 (4 x  1) dx 
4x 2
 x  c  2x 2  x  c
2
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BQT 133
Both answer are actually the same, because
1
1
(4 x  1) 2  c  (16 x 2  8 x  1)  c
8
8
1
 2x 2  x   c
8
 2x 2  x  C,
(b) If u  4x  1, then
1
c
8
du
1
 4  du  dx . Substituting u and du give us
dx
4
 (4 x  1)
Remarks
where C 
2
1 2
u du
4
1
 u3  c
12
1
 (4 x  1) 3  c
12
dx 
By integrating directly will get
2
 (4 x  1) dx   (16 x  8x  1) dx 
16 x 3
 4x 2  x  C
3
Which is equivalent to the previous because
1
1
(4 x  1) 3  c  (64 x 3  48 x 2  12 x  1)  c
12
12
16 3
1
x  4x 2  x   c
3
12
16
 x 3  4x 2  x  C,
3

where C 
1
c
12
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BQT 133
Example 4.9
Evaluate the following integrals.
c
(a) 
(b)  e 4 x dx
dx
ax  b
5
2
(c)  (10 x  9) dx
(d)  35 x dx
(e)
1
 4x  3
dx
Solutions:
du
1
 a  du  dx
dx
a
c
1 1
 ax  b dx  c  u  a du
c 1
  du
a u
c
c
 ln u  c  ln ax  b  c
a
a
(a) If u  ax  b , then
(b) If u  4 x , then
e
4x
du
1
 4  du  dx
dx
4
1
dx   e u  du
4
1 u

e du
4
1
 eu  c
4
1
 e4x  c
4

du
1
 10  du  dx
dx
10
5
5
1
 (10 x  9) 2 dx   u 2  10 du
5
1

u 2 du
10
(c) If u  10x  9 , then

5
1
1 u2
 
c
10 5
1
2
7
1 2
 u c
35
7
1
 (10 x  9) 2  c
35
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BQT 133
du
1
 5  du  dx
dx
5
1
dx   3u  du
5
1 u

3 du
5
1 3u
 
c
5 ln 3
35 x

c
5 ln 3
(d) If u  5x , then
3
5x

du
1
 4  du  dx
dx
4
1
1 1
 4 x  3 dx   u  4du
1 1
1
ln( 4 x  3)

du  ln u  c 
c
4 u
4
4
(e) If u  4x  3 , then

Exercise 4.9:
(a)  e 3 x 5 dx
(b)  sec 2 (3x  1) dx
(c)
 cos(1  4x) dx
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(ii) Integral of the form
BQT 133

f ' ( x)
dx
f ( x)
2x  3
dx . This is not one of our standard integrals, so how
 3x  5
shall we tackle it? This is an example of a type of integral which is very easy to deal with but
which depends largely on how keen your wits are. You will notice that if we differentiate the
denominator, we obtain the expression in the numerator. So, let u stand for the denominator,
i.e. u  x 2  3x  5 .
du

 2x  3
 du  (2 x  3)dx
dx
Consider the integral
x
2
The given integral can then be written in terms of u:
2x  3
1
dx 
du  ln u  c
2
u
x  3x  5


If we now put back what u stands for in terms of x, we get
2x  3
dx  ln( x 2  3x  5)  c
2
x  3x  5

Example 4.10
Evaluate the following indefinite integrals by using the substitution method.
2x  3
x
12 x 2  16
(a)  2
(b)  2
(c)  3
dx
dx
dx
x  3x  2
x 4
x  4x
Solutions:
(a) If u  x 2  3x  2 , then
x
2
du
 2 x  3  du  (2 x  3)dx
dx
2x  3
1
dx    du  ln u  c  ln( x 2  3x  2)  c
u
 3x  2
du
1
 2 x  du  xdx
dx
2
x
1 1
 x 2  4 dx   u  2 du
1 1
1
  du  ln u  c
2 u
2
1
 ln( x 2  4)  c
2
(b) If u  x 2  4 , then
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BQT 133
du
 3x 2  4  du  (3x 2  4)dx
dx
2
12 x  16
4(3x 2  4)
 x 3  4x dx   x 3  4x dx
1
 4  du
u
 4 ln u  c
(c) If u  x 3  4 x , then
 4 ln( x 3  4 x)  c
Exercise 4.10:
(a)
x
2
x3
dx
 6x  2
(iii)Integral of the form
 f ( x) f ' ( x) dx
Example 4.11
Evaluate the following integrals.

(e) 2 x( x  2)


dx
(d) 
x ln x
(a)
4 x(2 x 2  3) 6 dx
(b)
e x (e x  1) 3 dx (c)
5

ln x
dx
x
dx
Solutions:
(a) If u  2 x 2  3 , then
 4 x( 2 x
2
du
 4 x  du  4 x dx
dx

 3) 6 dx  u 6 du
u7
1
  c  (2 x 2  3)7  c
7
7
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BQT 133
(b) If u  e x  1 , then
 e (e
x
x
du
 e x  du  e x dx
dx

 1) 3 dx  u 3 du
u4

c
4
1
 (e x  1) 4  c
4
du 1
1
  du 
dx
dx x
x
ln x
1
dx  ln x  dx
x
x
(c) If u  ln x , then


  u du
u2

c
2
1
 (ln x) 2  c
2
du 1
1
  du 
dx
dx x
x
dx
1 1

 dx
x ln x
ln x x
1

 du
u
 ln u  c
 ln(ln x)  c
(d) If u  ln x , then



(e) If u  x  2 , then
 2x( x  2)
5
du
 1  du  dx . Also x  u  2
dx

  (2u  4)u du
  (2u  4u )du
dx  2(u  2)u 5 du
5
6
5
2u 7 4u 6
2u 7 2u 6

c 

c
7
6
7
3
2u 6
2( x  2) 6
3( x  2)  7  c

(3u  7)  c 
21
21

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BQT 133
Exercise 4.11:
Evaluate the following integrals.
(a)
4.6

2
xe x dx
(b)
x
3x  2 dx
APPLICATIONS OF INTEGRATION
4.6.1 Consumer’s Surplus and Supplier’s Surplus
Suppose that there are eight people in a store wanting to buy a can of chili and that the price of
the chili is not marked. Each person pays as much as he or she is willing to pay. One person pays
RM 6. Another pays RM5.50. The third pays RM5, the forth RM4.50, the fifth RM4, the sixth
RM3, the seventh RM2 and the eighth RM1. The store collects 6+5.50+5+4.50+4+3+2+1 =
RM31. If the store had marked a price of RM1, then all eight would have paid RM1 and the store
would have only collected RM8, a reduction of RM23. In other words, sales worth RM31 to the
consumers would have been purchased for RM8. The difference is called the consumer surplus. It
is the difference between what consumers are willing to pay for a product or service and what
they actually do pay. The consumer surplus represents the total savings to consumers who are
willing to pay more than price for the product. The supplier’s surplus measures the difference
between the amount of money a supplier is willing to accept at a given price for a product and the
amount the supplier actually does receive.
Definition 4.3
The consumer’s surplus at a price level of p is
q
Consumer surplus (CS)   D  x  dx  pq
0
Where D  x  is demand function and pq is the revenue.
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BQT 133
Definition 5.4
The supplier’s surplus at a price level of p is
q
Supplier surplus (SS)  pq   S  x  dx
Where S  x  is supply function and pq is the revenue.
0
Example 4.12
Find the consumer’s surplus for the given function and quantity
(a) D( x)  0.01x  x  20, q  15 .Answer:RM135
2
(b) D( x) 
20 x  40
, q  12 .Answer:RM19.70
x2  4 x  5
Example 4.13
2
Find the supplier’s surplus for S ( x)  0.05 x  x  10, q  20 . Answer RM466.67.
Example 414
Find the equilibrium point, consumer surplus and supplier surplus for the given supply and
demand function.
D  x    x  107.5 and S  x   0.02 x 2  2 x  20
Answers
Equilibrium point=25
CS=RM312.50
SS=RM833.33
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