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Integration
Integration is the reverse of differentiation
...
Given

f / x   4 , then f x   4 x  c
...

Solution:
The derivative of any constant function is zero
...

Example 2
Find the antiderivative of f / x   2 x  7
...
r
...

dx

Moreover, if c is any constant, then





d 5
x  c  5x 4
...


NOTE: Different values of c will give different integrals and thus integral of a function is unique
...


Rules of Integration
Rule 1
...


Example
Evaluate the following integrals

  7dx

(i)

1

 3 dx
 3 dx

(ii)
(iii)

10

  7 dx
 0dx

(iv)
(v)

3
dx
2



(vi)

Solution:

  7dx  7 x  c

(i)

1

1

 3 dx  3 x  c
 3 dx  3x  c

(ii)
(iii)

10

10

  7 dx   7 x  c
 0dx  0 x  c  c

(iv)
(v)



(vi)

3
3
dx 
xc
2
2

Rule 2: Power Rule

x n 1
 x dx  n  1  c , provided that n  1
...

Example
Evaluate the following integrals
(i)
(ii)
(iii)

 4 xdx
x2
 5 dx
 3 xdx

(iv)
(v)
(vi)
(vii)

5
dx
x



 20 x dx
 5 xdx
5

3



10
5

x4

dx

Solution:

4x2
 c  2x2  c
2
x2
x3
x3
 5 dx  5  3  c  15  c

 4xdx 

(i)
(ii)

3

(iii)

3
3
3x 2
2
3
x
dx

 c   3x 2  c  2 x 2  c

3
3
2

1

(iv)

(v)
(vi)

(vii)

1
1
1
5
5x 2
2
2
2
2
dx

5
x
dx


c


5
x

c

10
x
c
 x

1
1
2
20 6
10 6
5
 20 x dx  6 x  c  3 x  c
1
1 1
4
5
5 43
3
15 4 3
3
3
3
3
5
x
dx

5
x
dx

x

c

x

c


5
x

c

x c


1 1
4
4
4
3
3



10
5

x4

dx  10 x

4

1

5

1
10 x 5
dx 
 c  50 x 5  c
1
5

Rule 4: Addition or Difference
If

 f x dx

 g x dx exist, then   f x   g x dx   f x dx   g x dx

and

Example
Evaluate the following integrals
...


 x

(ii)
(iii)
(iv)



x3 5 x 2

 4x  c
3
2
2 7 2
 5 2

6
  6 x  x 4  7 x  5 dx  x  3x3  2 x  5x  c
1
2 3
3 2 9 2


  x  3 x  9 x  13dx  3 x 2  2 x 3  2 x  13x  c
2

 5x  4 dx 

 x 4  3x3  x 
x4
3
2


dx   x  3x  1 dx 
 x3  x  c
 

x
4




(v)

Rule 5: Power Rule Exception

1

 xdx   x

1

dx  ln x  c

Rule 6: Exponential Function
ax
 e dx 

eax
 c , where a is a constant
...

(i)
(ii)
(iii)
(iv)

 e dx
 e dx
 4e dx
 e dx
x

x

3x

x2 4x



1
7
 dx
x x

(v)

  2 x

(vi)

 2 x 3  3x 2  4 x  5 
dx
 
x2


(vii)

3

 4 x 3  3x 

 2  3x 1  4 x 7  2 x dx



2

(viii)

 2 1
  x  x 2  dx

Solution:
(i)
(ii)
(iii)
(iv)

ex
x
 e dx  1  c  e  c
e x
x
e
dx

 c  e x  c

1
4 3x
3x
 4e dx  3 e  c
2
e x 4 x
 x2 4 x
 e dx   2 x  4  c
x

1

(v)

(vi)

(vii)

1 7
2 4 4 2 3 2 x 2
 3
3
2
x

4
x

3
x


dx

x 
x  x 
 7 ln x  c



1
4
2
2
x x
2
1
1
3
 x 4  2 x 2  x 2  2 x 2  7 ln x  c
2
2
 2 x 3  3x 2  4 x  5 
 2 x 3 3x 2 4 x 5 


dx

 
  x2  x2  x2  x2 dx

x2


4 5

   2 x  3   2 dx
x x 

2x2
5

 3x  4 ln x   c
2
x
3
2
 2  3x 1  4 x 7  2 x dx   24 x  94 x  31x  14 dx





24 4 94 3 31 2
x  x  x  14 x  c
4
3
2
94
31
 6 x 4  x 3  x 2  14 x  c
3
2


2

(viii)

x 5 2 3 1 3
 2 1
 4 2 1
x

dx

x


dx





  x2 
  x4 x4  5  3 x  3 x  c

Exercise
Evaluate the following integrals
...
Integration by Substitution
...

d 2
2x  7
For example, to evaluate  2
x  7 x  13  2 x  7
...
However, the following suggestions will be useful
...

a
du 1
f u 
f ax  b 
Thus  f / ax  b dx   f / u 
  f / u du 
c 
c
a
a
a
a
(ii)
When the integrand is of the form x n1 f / x n
...

dx
du 1
1
1
Thus  x n 1 f / x n dx   f / u 
  f / u du  f u   c  f x n  c
n
n
n
n
n
/
(iii)
When the integrand is of the form  f  x 
...





2



 

 

 

We put u  f x  and du  f

  f x 

n

f

/

x    u n du 

We put u  f x  and du  f

f x 
/

Thus
(v)

 f x dx  

/

xdx
...

We put u  f x  and du  f
Thus

Example

xdx

 f x   c
u n 1
c 
n 1
n 1
/
f x 
When the integrand is of the form

...

(i)

x

(ii)



(iii)



2

2x  3
dx
 3x  7
dx

xx
x2
3x 3  7

5

dx

(viii)

 x 1  x  dx
 2 xe dx
 x e dx
 x x  5 dx
 2x 1  x dx

(ix)



(x)

e

(iv)
(v)
(vi)
(vii)

4

1

5

3

x2

2 3 x3

2

3

2

1
2x  1
x

dx

1  e x dx

Solution:
(i)

x

2

2x  3
dx
 3x  7

Let u  x 2  3x  7 , then

2x  3
du
dx  
 ln u  c  ln x 2  3x  7  c
u
 3x  7
dx
dx
 x  x   x 1 x
dx
du
1
Let u  1  x , then
or 2du 

dx 2 x
x
du
2du
1
 x 1  x   u  2 u du  2 ln u  c  2 ln 1  x  c

x

(ii)



2





(iii)

du
 2 x  3 or du  2 x  3dx
dx



x2
3x 3  7







dx  



3x

x2



1

dx



7
du
du
Let u  3 x 3  7 , then
 9x 2 or du  9 x 2 dx or
 x 2 dx
dx
9
5

3

5



x2
5

3x 3  7

dx  

3x

x2
7

3



1

dx  

5

du
9u

1

5

1 1
1 15
1  u 5 
  u du 
c
9
9   1  1
 5 

1  u 5 
 
c
94 
 5
1 5 4
  u 5
9 4
5


3x 3  7   c
36
4

(iv)

 x 1  x 
4

5

1

3

dx

du
du
 x 4 dx
 5x 4 or
dx
5
4
1
4
1 du
1 13
1  u 3 
3 43
3
4
5 3
5 3
3




x
1

x
dx

u

u
du


c

u

c

1

x
c

 5 5
54 
20
20
 3

Let u  1  x 5 , then

(v)

 2 xe

x2

dx

du
 2 x or du  2 xdx
dx
eu
x2
u
x2
2
xe
dx

e
du


c

e
c


1

Let u  x 2 , then

(vi)

x

2 3 x3

e

dx

du
du
 x 2 dx
 9x 2 or
dx
9
3
1
1
1 3 x3
2 3x
u
u
 x e dx   9e du  9 e  c  9 e  c

Let u  3x 3 , then

(vii)



2
3
2
3
 x x  5 dx   x x  5

Let u  x 3  5 , then

x
(viii)

2



1

2

dx

du
du
 3x 2 or
 x 2 dx
dx
3

1
 u 32 
2
3
u
1
  c  2 u 2  c  2 x3  5
x  5 dx   x x  5 2 dx  
du  
3
3 3 
9
9
 2
3

2





2
2
 2x 1  x dx   2x 1  x

Let u  1  x 2 , then



3



1



1

2

dx

du
 2 x or du  2 xdx
dx



2
2
 2 x 1  x dx   2 x 1  x



1

2

1

dx   u 2 du 



2 32
2
u  c  1 x2
3
3



3

2

c



3

2

c

(ix)

1



2x  1

dx  

1

2 x  1 12

Let u  2 x  1 , then

dx

du
du
 dx
 2 or
dx
2

1
1
1 du 1 12
1  u 2 
dx  
dx   1
  u du 
c u 2 c
1
2
2 1 
2x  1
2 x  1 2
u 2 2
 2

1



1

 2 x  1 2  c
1

(x)

e



Let u  1  e x , then

du
 e x or du  e x dx
dx

e



1  e x dx   e x 1  e x

x

Exercise
Evaluate
ln x
dx
x

(i)



(ii)

 4 x e dx
 
 2 x  1e dx
4
2
 xx  3 dx
2
 x 1  x dx

(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)



1  e x dx   e x 1  e x dx

x

3

x4

x2  x

3

 x 1  x dx
 x x  2 dx
 e 1  e  dx
 xe dx
3

2

2

4

x 3

x

x 2 1

 x

x2

dx
4
8
dx
 e x  1dx
1
 x1  ln xdx

 e

3

x



 e x

 e
2

x



 e  x dx



1

2

1

dx   u 2 du 



2 32
2
u  c  1 ex
3
3



3

2

c

2
...

If u and v are two functions of x , then
 du

 uvdx  u  vdx    dx  vdxdx
In words, integral of the product of two functions
...

(ii)
In case of integrals of the form  f x 
...


(iii)

In case of integrals of the form

ln x n

(iv)

 ln x  dx
n

as the first function
...


Example 1
Evaluate the following
...
e
...
e
...
e
...
e
...
(1)

Again applying integration by parts for  2 ln xdx

 udv  uv   vdu
Let u  ln x , then
we get

du 1
dx
or du 
and let dv  2dx , on integrating both sides

dx x
x

v  2x

 2 ln xdx  2 x ln x   2 x

dx
x

 2 x ln x   2dx

 2 x ln x  2 x  c …………
...
 e x  1  c

Ans
...
e x x 2  2 x  2  c





Ans
...

ln x   c
3
9
Ans
...
x  1 ln1  x   x  c

3
...

Example
Evaluate the following

x

(i)

 x  12x  1dx

(ii)

x

(iii)
(iv)

2x  3
dx
 2x  3
1
 x 2  x  6dx
7 x 2  3x  1
 x 2  x dx
2

Solution:
(i)

(ii)

x
A
B


x  12 x  1 x  1 2 x  1
A2 x  1  Bx  1

2 x  1x  1
 x  A2 x  1  Bx  1
1
Putting x  1 , we get A 
3
1
1
Putting x 
, we get B 
2
3
x
1 dx 1 dx
Thus 
dx  

x  12 x  1
3 x  1 3  2x  1
1
1
 lnx  1  ln2 x  1  c
3
3
2x  3
2x  3
A
B
Let 2



x  2 x  3 x  3x  1 x  3 x  1
Ax  1  Bx  3

x  3x  1
 2 x  3  Ax  1  Bx  3
9
Putting x  1 , we get B 
4
Let

1
4
2x  3
9 dx
1 dx
 x 2  2 x  3dx  4  x  3  4  x  1
9
1
 lnx  3  lnx  1  c
4
4

Putting x  3 , we get A 

(iii)

(iv)

1
1
A
B



x  x  6 x  3x  2 x  3 x  2
A x  2   B x  3

x  3x  2
 1  Ax  2  Bx  3
1
Putting x  2 , we get B  
...

Finding the area between a curve and the x  axis
...

Solution:

Area  

4

1

4



x3 
4 3   2 13 
2
6 x  x dx  3x 2    34    31    48  24  24
3 1 
3 
3




2



Finding the area between two curves
...

2

x


   x 2   2 dx
2

 2
1

1

 x3 x2

 
 2 x
4
3
 2


13 12
   23  22
    21  

 2 2
4
3 4
  3

1 1
  8 4

    2  
  4   8
...

Example
Evaluate

 x
1

(i)

2



 1 dx

0

2

(ii)



6 x  4dx

0

4

1

(iii)

 xdx

(iv)

 e

2

2

1

Solution:

3x



 3x 2 dx

(i)

0

1

 x3

x  1 dx    x 
3
0


1

2



13   0 3

   1    0
3   3

1
 1 0  0
3
4

3
2

(ii)



6 x  4dx

0

Let u  6 x  4 , then
2

2



6 x  4dx  

0

0

du
du
 6 or dx 
dx
6

  32
du 1
1 u
u
  u du   
6  3
6 60
  2
2

1
2

2

2
2

   1  2 u 3 2    1 u 3 2 
 6 3

 9


0
0
 0
2

3 
1
  6 x  4 2 
9
0
3 
3 
1
1
  6  2  4 2    6  0  4 2 
9
 9

3
1
1 3
 16 2  4 2
9
9
56

9
4

(iii)

 xdx  ln x

(iv)



1

4
2

 ln 4  ln 2  ln 2 2  ln 2  2 ln 2  ln 2  ln 2

2

2

1

2

 e3x

e6
  e3  e6
e3
1
e  3x dx  
 x 3     8    1 
 8   1  e6  e3  7
3
3
 3
1  3
 3
 3
3x

2







Exercise
1
...


2
...


3
...

2
5
...

6
...
Find the area bounded by f x   x 2  1 , g x  

3

(i)

 x

x2 1
 3x

3

2

3

(ii)

 2 x
2

0

3

dx

 3x

2

dx


 2 x  1dx

 2x  2

2

(iv)



 x 1
x 1

4

 x

2

8x  2
2

1

(iii)

dx



3

0

(v)

 e
1

0

x



2

 e  x dx

2

(vi)

x

3 x  2dx

1

(vii)
(viii)

1 1 x
e dx
1 x2
3
2
1 3x  4 x  1
dx
0
x



2

Applications of Integration to cost, revenue and profit functions
...
e
...
Since integration is the inverse of differentiation,
dx
therefore, the total cost function is the integral of the marginal cost function,
i
...
C x    MCdx  k
where k is the arbitrary constant of integration
...
e
...


Revenue function
The total revenue function Rx  is given, then the marginal revenue function MR , is the
dR
derivative of the total revenue function i
...
, MR 

...
e
...

Also we know that Rx   px , where p is the price  p 

Rx 
, which is the demand function
...
The marginal cost function of manufacturing x shoes is 6  10 x  6 x 2
...
Find the total and average cost function
...

Also we are given when x  2 , then C x   12
12  6  2  10 

4 6 3
 2 k
2 3

12  12  20  16  k

k  4

Hence the total cost function Cx   6 x  5x 2  2 x 3  4
AC, by definition, is

C x 
4
 6  5x  2 x 2 
x
x

2
...

Solution:
The marginal cost is

MC  22 x  9

1

2

1

C x    22 x  9 2 dx
Let u  2 x  9 , then
1

du
du
 2 or dx 
dx
2

C x    22 x  9 dx   2u
2

1

1

2

1
1
du
u 2
  u 2 du 
 k  2u 2  k
1
2
2

 Cx   22 x  9 2  k
1

Where k is the constant of integration
...
A manufacturer’ s marginal cost function is MC  0
...
6 x  40 , where x is the
number of units of a product
...

Solution:
MC  0
...
0
...
003x



200



2

 0
...
003  0
...
Marginal cost of production in dollars is 3 
 e 0
...
Find the cost of producing 100 units
...
05 (approximately)
...
03 x
3000
Total cost of producing 100 units is



100

100

0

0

x

 MCdx    3  3000  e

0
...
03 x 
 3x 


6000 0
...
03  0
...
03
100
 300 
 333
...
05 ]

5
...
It is also known that total cost equals Ksh
...
Determine the total cost function
...

100 2
40,000 
 100100  c
2
40,000  5,000  10,000  c
c  25,000
The specific function representing the total of producing the product is
x2
 C x  
 100x  25,000
2
6
...
Campaign expenditures will be incurred at a rate of £
10,000 per day
...
The
function describing the rate at which contribution are received is C t   100t 2  20,000
where t represents the day of the campaign , and C t  equals the rate at which
contributions are received, measured in shillings per day
...

(i)
Determine how long the campaign should be conducted in order to maximize
net proceeds
...
67
 100

Netproceeds  Costs  Expenditures
 166,666
...
67
7
...
Acting as consultants, they have estimated the demand curve of a client’s firm to
be AR  200  8Q where AR is the average revenue in shillings and Q is the output in
units
...
Investigations have shown
that the firm’s cost when not producing output is Ksh
...

Required:
(i)
The total cost function
(ii)
The total revenue function
(iii)
The level of output that maximize profit

(iv)

Solution:
(i)

The total cost function





1
TC   MCdq   q 2  28q  211 dq  q 3  14q 2  211q  c
3
1 3
 C x   q  14q 2  211q  c
3
1 3
2
Given that C 0  10 , then C 0  0  140  2110  c
3

10  c

(ii)
(iii)

1
 C x   q 3  14q 2  211q  10
3
The total revenue function
TR  AR  Q  200  8q q  200q  8q 2
The level of output that maximizes profit
1
Pr ofit  TR  TC  200q  8q 2  ( q 3  14q 2  211q  10)
3
1
 200q  8q 2  q 3  14q 2  211q  10
3
1 3
P   q  6q 2  11q  10
3
d 2P
dP
0
 0 and
To maximize profit,
dq
dq 2

dP
 q 2  12q  11  0
dq
 12  12 2  4  1  11  12  10
q 

2
2
 12  10
 12  10
q
 11
 1 or q 
2
2
d 2P
 2q  12
dq 2
d 2P
 21  12  10  0 hence minimum
...

dq 2
 the level of output that will maximize profit is 11 units
...
If the marginal revenue is given by 15  2 x  x 2 , find the total revenue and demand
function
...

Solution:
d
We know that MR  TR 
dx
 TR   MRdx  k
When q  11 ,

Thus, TR   15  2 x  x 2 dx  k

Rx   15x  x 2 

x3
k
3

At x  0 , R0  0 and thus k  0
...

Also then, maximum revenue
27
 15  3  9 
 27
3
9
...
If the total revenue equals 0 when no unit are sold,
determine the total revenue function for the producer
...

2
10
...

(i)
What is the expected maintenance expenditures during the automobile first 5
years?
(ii)
Of the above expenditures, what is expected to be incurred during the fifth
year?

Solution:
3

10t 3 
105


100

10
t
dt

100
t


100
5

 500  416
...
67


0
3 0
3

Of these expenditures, those expected to be incurred during the fifth year are
estimated as
5

(i)
(ii)



2

5



5


10t 3 
100

10
t
dt

100
t



4
3 4

5



2



3
3

105  
104 
 1005 
  1004 

3  
3 


 916
...
33
 303
...
If the marginal revenue function for a commodity is MR  9  4 x 2 , find the demand
function
...
The marginal revenue (in thousands of dollars) function for a particular commodity is
5  3e 0
...
Determine the total revenue from
the sale of 100 units
...
05 approximately
...
03 x
Rx    MRdx  k

If p is the price Rx   px and so the demand function is p 

 5  3e

100



0

0
...
03 x 
 5 x  3
 0
...
03  
0
...
05  100
 595
 59500

13
...
3x 2
...

Solution:
The marginal revenue is
dR
 275  x  0
...
e
...
3x 2 i
...


20

Now

R20  R10   MRdx
10

 275  x  0
...
3 
2
3 10


3
2



20 2
20  
10 2
10 
 275  20 
 0
...
3
  275  10 

2
3  
2
3 

 1900

Exercise
1
...

6
 5 , find the total
2
...

3
...
0002x 2 , where

R / denotes marginal revenue and x denotes the quantity produced and sold
...
Find the market demand function for the
commodity
...
A company is specializing in mail order sales approach is beginning a promotional
campaign
...
Marketing
specialists estimates that the rate at which profit (exclusive of advertising costs) will be
generated from the promotion campaign decreases over the length of the campaign
...


6
...

8
...
In
order to maximize net profit, the firm should conduct the campaign as long as r t 
exceeds the daily advertising cost
...

(ii)
How long should the campaign be conducted?
(iii)
What are the total advertising expenditures expected to equal during the
campaign?
(iv)
What net profit will be expected?
dR
2000

A manufacturer’s marginal revenue function is:

...

The marginal cost is given by MC  150  e5 x , where x is the number of units, find the
average cost function
...
01x  0
...
Find the total
cost function and the average cost function if the cost of producing 10 units is 105 dollars
...

(i)
Find c , if C 0  100
...

(iii)
Evaluate c , MC and AC for x  60
...
If the marginal cost function is given by MC  33x  4 2 and fixed cost is 2 dollars, find
the average cost for 7 units of output
...
The marginal cost of production is found to be MC  1000  20 x  x 2 , where x is the
number of units produced
...
Find the cost
function
...
If the marginal revenue function for output x is given by MR 
 1
...

the average revenue function is equal to p 
6x  9
1



---