ICAI Notes- Integral Calculus | Quantitative Aptitude for CA Foundation PDF Download

 Page 1


8. 23 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
Integration is the reverse process of differentiation.
Integration
f(x) f’(x)
Differentiation
We know
       
? ?
?? 1n
x 1n
1n
x
dx
d
n 1n
?
?
?
?
?
?
?
?
?
?
?
?
?
or 
n
1n
x
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
?
            
…………………(1)
Integration is the inverse operation of differentiation and is denoted by the symbol 
?
.
Hence, from equation (1), it follows that
n+1
n
x
x dx
n+1
?
?
i.e. Integral of x
n
 with respect to variable x is equal to 
n+1
x
n+1
Thus if we differentiate 
? ?
n+1
x
n+1
 we can get back x
n
Again if we differentiate 
? ?
n+1
x
n+1
 + c and c being a constant, we get back the same x
n
 .
i.e. 
n
1n
xc
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
Hence  
?
x
n
 dx = 
? ?
n+1
x
n+1
 + c and this c is called the constant of integration.
Integral calculus was primarily invented to determine the area bounded by the curves dividing
the entire area into infinite number of infinitesimal small areas and taking the sum of all these
small areas.
© The Institute of Chartered Accountants of India
Page 2


8. 23 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
Integration is the reverse process of differentiation.
Integration
f(x) f’(x)
Differentiation
We know
       
? ?
?? 1n
x 1n
1n
x
dx
d
n 1n
?
?
?
?
?
?
?
?
?
?
?
?
?
or 
n
1n
x
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
?
            
…………………(1)
Integration is the inverse operation of differentiation and is denoted by the symbol 
?
.
Hence, from equation (1), it follows that
n+1
n
x
x dx
n+1
?
?
i.e. Integral of x
n
 with respect to variable x is equal to 
n+1
x
n+1
Thus if we differentiate 
? ?
n+1
x
n+1
 we can get back x
n
Again if we differentiate 
? ?
n+1
x
n+1
 + c and c being a constant, we get back the same x
n
 .
i.e. 
n
1n
xc
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
Hence  
?
x
n
 dx = 
? ?
n+1
x
n+1
 + c and this c is called the constant of integration.
Integral calculus was primarily invented to determine the area bounded by the curves dividing
the entire area into infinite number of infinitesimal small areas and taking the sum of all these
small areas.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 24
i)
1 n
x
dx x
1n
n
?
?
?
?
 + c,
? ? ?
n+1
x1
n 1 (If n=-1, which is not defined)
n+1 0
ii) ? dx  = x + c, since ?  1dx = ? x°dx=
1
1 x
 = x + c
iii)
?
e
x
 dx = e
x
 + c, since 
xx
d
e =e
dx
iv)
?
e
ax
 dx = c
a
ax
e
? , since 
ax
e
a
ax
e
dx
d
?
?
?
?
?
?
?
?
?
v)
?
dx
x
= log x+c, since 
d 1
logx=
dx x
vi)
?
a
x
 dx = a
x
 / log
e
a+c, since 
d a
a
dx log
x
x
a
e
??
? ?
? ? ?
?
?
? ?
??
Note: In the answer for all integral sums we add +c (constant of integration) since the differentiation of
constant is always zero.
Elementary Rules:
?
c f(x) dx = c 
?
f(x) dx where c is constant.
?
 { f(x) dx ± g(x)} dx = 
?
f(x)dx ± 
?
g(x)dx
Examples : Find (a) 
?
, dxx
(b)
,dx
x
1
?
(c) 
?
?
dxe
x3
   (d) 
?
dx3
x
 (e)
?
. dx x x
Solution: (a) ?
x
 dx = x
1/2 +1
 / (1/2 + 1)   = 
3/2
 x
3/2
   = 
?
3/2
2
3
x
c
(b) 
?
dx
x
1
 = 
c x 2c
1
2
1
x
dxx
1
2
1
2
1
? ? ?
? ?
?
? ?
?
?
 where c is arbitrary constant.
(c)  
?
?
dx e
x 3
  = c e
3
1
c
3
e
x3
x3
? ?? ?
?
?
?
(d) 
?
dx
x
3 =
.c
3log
3
e
x
?
© The Institute of Chartered Accountants of India
Page 3


8. 23 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
Integration is the reverse process of differentiation.
Integration
f(x) f’(x)
Differentiation
We know
       
? ?
?? 1n
x 1n
1n
x
dx
d
n 1n
?
?
?
?
?
?
?
?
?
?
?
?
?
or 
n
1n
x
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
?
            
…………………(1)
Integration is the inverse operation of differentiation and is denoted by the symbol 
?
.
Hence, from equation (1), it follows that
n+1
n
x
x dx
n+1
?
?
i.e. Integral of x
n
 with respect to variable x is equal to 
n+1
x
n+1
Thus if we differentiate 
? ?
n+1
x
n+1
 we can get back x
n
Again if we differentiate 
? ?
n+1
x
n+1
 + c and c being a constant, we get back the same x
n
 .
i.e. 
n
1n
xc
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
Hence  
?
x
n
 dx = 
? ?
n+1
x
n+1
 + c and this c is called the constant of integration.
Integral calculus was primarily invented to determine the area bounded by the curves dividing
the entire area into infinite number of infinitesimal small areas and taking the sum of all these
small areas.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 24
i)
1 n
x
dx x
1n
n
?
?
?
?
 + c,
? ? ?
n+1
x1
n 1 (If n=-1, which is not defined)
n+1 0
ii) ? dx  = x + c, since ?  1dx = ? x°dx=
1
1 x
 = x + c
iii)
?
e
x
 dx = e
x
 + c, since 
xx
d
e =e
dx
iv)
?
e
ax
 dx = c
a
ax
e
? , since 
ax
e
a
ax
e
dx
d
?
?
?
?
?
?
?
?
?
v)
?
dx
x
= log x+c, since 
d 1
logx=
dx x
vi)
?
a
x
 dx = a
x
 / log
e
a+c, since 
d a
a
dx log
x
x
a
e
??
? ?
? ? ?
?
?
? ?
??
Note: In the answer for all integral sums we add +c (constant of integration) since the differentiation of
constant is always zero.
Elementary Rules:
?
c f(x) dx = c 
?
f(x) dx where c is constant.
?
 { f(x) dx ± g(x)} dx = 
?
f(x)dx ± 
?
g(x)dx
Examples : Find (a) 
?
, dxx
(b)
,dx
x
1
?
(c) 
?
?
dxe
x3
   (d) 
?
dx3
x
 (e)
?
. dx x x
Solution: (a) ?
x
 dx = x
1/2 +1
 / (1/2 + 1)   = 
3/2
 x
3/2
   = 
?
3/2
2
3
x
c
(b) 
?
dx
x
1
 = 
c x 2c
1
2
1
x
dxx
1
2
1
2
1
? ? ?
? ?
?
? ?
?
?
 where c is arbitrary constant.
(c)  
?
?
dx e
x 3
  = c e
3
1
c
3
e
x3
x3
? ?? ?
?
?
?
(d) 
?
dx
x
3 =
.c
3log
3
e
x
?
© The Institute of Chartered Accountants of India
8 . 25 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
(e) 
?
. d x x x = 
3
1
3
2
5/2
2
2
.
3
5
1
2
x
x dx dx x c
?
? ? ?
?
?
Examples : Evaluate the following integral:
i)
?
(x + 1/x)
2
 dx = 
?
x
2
 dx +2 
?
dx + 
?
dx / x
2
= 
3 –2+1
x x
+2x+
3 –2+1
= 
3
x 1
+2x– +c
3 x
ii)
? x
 (x
3
+2x –3 ) dx = ?
?
x
7/2
 dx +2 
?
x
3/2
 dx –3 
?
x
1/2
 dx
= 
7/2+1 3/2+1 1/2+1
x 2 x 3 x
  +   –        
7/2 +1 3/2+1 1/2+1
= 
9/2 5/2
3/2
2x 4x
 +   –2 x + c 
9 5
iii)
?
 e
3 x
+ e
–4 x
 dx
= 
?
e
2 x
 dx      +  ?
?
e
–4 x
 dx
= 
4
e
2
e
x 4 x 2
?
?
?
  = 
c
e 4
1
2
e
x 4
x 2
? ?
iv)  
? ?
?
?
?
?
dx
1 x
1 1 – x
dx
1 x
x
2 2
                        =  
    
1 x
dx
  dx 
1 x
1) - (x
2
? ?
?
?
?
                         = 
?
? ? ? ? ? ? ? ? c ) 1 x log( x
2
x
) 1 x log( dx ) 1 x (
2
v)
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
By simple division   
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
= 
? ?
2
9
x 3x 6 dx
x 2
? ?
? ?
? ? ? ?
? ?
?
? ?
? ?
?
© The Institute of Chartered Accountants of India
Page 4


8. 23 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
Integration is the reverse process of differentiation.
Integration
f(x) f’(x)
Differentiation
We know
       
? ?
?? 1n
x 1n
1n
x
dx
d
n 1n
?
?
?
?
?
?
?
?
?
?
?
?
?
or 
n
1n
x
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
?
            
…………………(1)
Integration is the inverse operation of differentiation and is denoted by the symbol 
?
.
Hence, from equation (1), it follows that
n+1
n
x
x dx
n+1
?
?
i.e. Integral of x
n
 with respect to variable x is equal to 
n+1
x
n+1
Thus if we differentiate 
? ?
n+1
x
n+1
 we can get back x
n
Again if we differentiate 
? ?
n+1
x
n+1
 + c and c being a constant, we get back the same x
n
 .
i.e. 
n
1n
xc
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
Hence  
?
x
n
 dx = 
? ?
n+1
x
n+1
 + c and this c is called the constant of integration.
Integral calculus was primarily invented to determine the area bounded by the curves dividing
the entire area into infinite number of infinitesimal small areas and taking the sum of all these
small areas.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 24
i)
1 n
x
dx x
1n
n
?
?
?
?
 + c,
? ? ?
n+1
x1
n 1 (If n=-1, which is not defined)
n+1 0
ii) ? dx  = x + c, since ?  1dx = ? x°dx=
1
1 x
 = x + c
iii)
?
e
x
 dx = e
x
 + c, since 
xx
d
e =e
dx
iv)
?
e
ax
 dx = c
a
ax
e
? , since 
ax
e
a
ax
e
dx
d
?
?
?
?
?
?
?
?
?
v)
?
dx
x
= log x+c, since 
d 1
logx=
dx x
vi)
?
a
x
 dx = a
x
 / log
e
a+c, since 
d a
a
dx log
x
x
a
e
??
? ?
? ? ?
?
?
? ?
??
Note: In the answer for all integral sums we add +c (constant of integration) since the differentiation of
constant is always zero.
Elementary Rules:
?
c f(x) dx = c 
?
f(x) dx where c is constant.
?
 { f(x) dx ± g(x)} dx = 
?
f(x)dx ± 
?
g(x)dx
Examples : Find (a) 
?
, dxx
(b)
,dx
x
1
?
(c) 
?
?
dxe
x3
   (d) 
?
dx3
x
 (e)
?
. dx x x
Solution: (a) ?
x
 dx = x
1/2 +1
 / (1/2 + 1)   = 
3/2
 x
3/2
   = 
?
3/2
2
3
x
c
(b) 
?
dx
x
1
 = 
c x 2c
1
2
1
x
dxx
1
2
1
2
1
? ? ?
? ?
?
? ?
?
?
 where c is arbitrary constant.
(c)  
?
?
dx e
x 3
  = c e
3
1
c
3
e
x3
x3
? ?? ?
?
?
?
(d) 
?
dx
x
3 =
.c
3log
3
e
x
?
© The Institute of Chartered Accountants of India
8 . 25 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
(e) 
?
. d x x x = 
3
1
3
2
5/2
2
2
.
3
5
1
2
x
x dx dx x c
?
? ? ?
?
?
Examples : Evaluate the following integral:
i)
?
(x + 1/x)
2
 dx = 
?
x
2
 dx +2 
?
dx + 
?
dx / x
2
= 
3 –2+1
x x
+2x+
3 –2+1
= 
3
x 1
+2x– +c
3 x
ii)
? x
 (x
3
+2x –3 ) dx = ?
?
x
7/2
 dx +2 
?
x
3/2
 dx –3 
?
x
1/2
 dx
= 
7/2+1 3/2+1 1/2+1
x 2 x 3 x
  +   –        
7/2 +1 3/2+1 1/2+1
= 
9/2 5/2
3/2
2x 4x
 +   –2 x + c 
9 5
iii)
?
 e
3 x
+ e
–4 x
 dx
= 
?
e
2 x
 dx      +  ?
?
e
–4 x
 dx
= 
4
e
2
e
x 4 x 2
?
?
?
  = 
c
e 4
1
2
e
x 4
x 2
? ?
iv)  
? ?
?
?
?
?
dx
1 x
1 1 – x
dx
1 x
x
2 2
                        =  
    
1 x
dx
  dx 
1 x
1) - (x
2
? ?
?
?
?
                         = 
?
? ? ? ? ? ? ? ? c ) 1 x log( x
2
x
) 1 x log( dx ) 1 x (
2
v)
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
By simple division   
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
= 
? ?
2
9
x 3x 6 dx
x 2
? ?
? ?
? ? ? ?
? ?
?
? ?
? ?
?
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 26
=  
2
x3
3
x
23
?
 – 6x+9log(x+2)+c
It is sometime possible by a change of independent variable to transform a function into another
which can be readily integrated.
We can show the following rules.
To put z = f (x) and also adjust dz = f
 ? 
(x) dx
Example: 
, dx ) x( ' h } ) x( h { F
?
 take e
z
 = h(x) and to adjust dz = h
?
 (x) dx
then integrate 
?
dzz F )( using normal rule.
Example:  
?
? dx ) 3 (2x
7
We put (2x + 3) = t   ?   so 2 dx = dt    or  dx = dt / 2
        Therefore 
? ?
? ?
?
?
? ? ? ? ?      c
16
3x 2
16
t
      
2x8
t
    dt t ½  dx ) 3 (2x
8 8 8
7 7
This method is known as Method of Substitution
Example:  
??
?
?
dx
1x
x
3
2
3
We put  (x
2
 +1) = t
so  2x dx = dt    or  x dx = dt / 2
=   
?
dx
t
x .x
3
2
=   
?
?
dt
t
1 t
2
1
3
=   
? ? 3 2
t
dt
2
1
–
t
dt
2
1
=   
?? ) 13 (–
t
2
1
–
12 –
t
2
1
13 – 12 –
?
?
?
?
? ?
= – 
2
1
  
t
1
 + 
4
1
 
2
t
1
=  
t
1
2
1
t
1
4
1
2
?
© The Institute of Chartered Accountants of India
Page 5


8. 23 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
Integration is the reverse process of differentiation.
Integration
f(x) f’(x)
Differentiation
We know
       
? ?
?? 1n
x 1n
1n
x
dx
d
n 1n
?
?
?
?
?
?
?
?
?
?
?
?
?
or 
n
1n
x
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
?
            
…………………(1)
Integration is the inverse operation of differentiation and is denoted by the symbol 
?
.
Hence, from equation (1), it follows that
n+1
n
x
x dx
n+1
?
?
i.e. Integral of x
n
 with respect to variable x is equal to 
n+1
x
n+1
Thus if we differentiate 
? ?
n+1
x
n+1
 we can get back x
n
Again if we differentiate 
? ?
n+1
x
n+1
 + c and c being a constant, we get back the same x
n
 .
i.e. 
n
1n
xc
1n
x
dx
d
?
?
?
?
?
?
?
?
?
?
Hence  
?
x
n
 dx = 
? ?
n+1
x
n+1
 + c and this c is called the constant of integration.
Integral calculus was primarily invented to determine the area bounded by the curves dividing
the entire area into infinite number of infinitesimal small areas and taking the sum of all these
small areas.
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 24
i)
1 n
x
dx x
1n
n
?
?
?
?
 + c,
? ? ?
n+1
x1
n 1 (If n=-1, which is not defined)
n+1 0
ii) ? dx  = x + c, since ?  1dx = ? x°dx=
1
1 x
 = x + c
iii)
?
e
x
 dx = e
x
 + c, since 
xx
d
e =e
dx
iv)
?
e
ax
 dx = c
a
ax
e
? , since 
ax
e
a
ax
e
dx
d
?
?
?
?
?
?
?
?
?
v)
?
dx
x
= log x+c, since 
d 1
logx=
dx x
vi)
?
a
x
 dx = a
x
 / log
e
a+c, since 
d a
a
dx log
x
x
a
e
??
? ?
? ? ?
?
?
? ?
??
Note: In the answer for all integral sums we add +c (constant of integration) since the differentiation of
constant is always zero.
Elementary Rules:
?
c f(x) dx = c 
?
f(x) dx where c is constant.
?
 { f(x) dx ± g(x)} dx = 
?
f(x)dx ± 
?
g(x)dx
Examples : Find (a) 
?
, dxx
(b)
,dx
x
1
?
(c) 
?
?
dxe
x3
   (d) 
?
dx3
x
 (e)
?
. dx x x
Solution: (a) ?
x
 dx = x
1/2 +1
 / (1/2 + 1)   = 
3/2
 x
3/2
   = 
?
3/2
2
3
x
c
(b) 
?
dx
x
1
 = 
c x 2c
1
2
1
x
dxx
1
2
1
2
1
? ? ?
? ?
?
? ?
?
?
 where c is arbitrary constant.
(c)  
?
?
dx e
x 3
  = c e
3
1
c
3
e
x3
x3
? ?? ?
?
?
?
(d) 
?
dx
x
3 =
.c
3log
3
e
x
?
© The Institute of Chartered Accountants of India
8 . 25 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
(e) 
?
. d x x x = 
3
1
3
2
5/2
2
2
.
3
5
1
2
x
x dx dx x c
?
? ? ?
?
?
Examples : Evaluate the following integral:
i)
?
(x + 1/x)
2
 dx = 
?
x
2
 dx +2 
?
dx + 
?
dx / x
2
= 
3 –2+1
x x
+2x+
3 –2+1
= 
3
x 1
+2x– +c
3 x
ii)
? x
 (x
3
+2x –3 ) dx = ?
?
x
7/2
 dx +2 
?
x
3/2
 dx –3 
?
x
1/2
 dx
= 
7/2+1 3/2+1 1/2+1
x 2 x 3 x
  +   –        
7/2 +1 3/2+1 1/2+1
= 
9/2 5/2
3/2
2x 4x
 +   –2 x + c 
9 5
iii)
?
 e
3 x
+ e
–4 x
 dx
= 
?
e
2 x
 dx      +  ?
?
e
–4 x
 dx
= 
4
e
2
e
x 4 x 2
?
?
?
  = 
c
e 4
1
2
e
x 4
x 2
? ?
iv)  
? ?
?
?
?
?
dx
1 x
1 1 – x
dx
1 x
x
2 2
                        =  
    
1 x
dx
  dx 
1 x
1) - (x
2
? ?
?
?
?
                         = 
?
? ? ? ? ? ? ? ? c ) 1 x log( x
2
x
) 1 x log( dx ) 1 x (
2
v)
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
By simple division   
          
    dx
2) (x
3 – 5x  x
2 3
?
?
?
= 
? ?
2
9
x 3x 6 dx
x 2
? ?
? ?
? ? ? ?
? ?
?
? ?
? ?
?
© The Institute of Chartered Accountants of India
BUSINESS MATHEMATICS
8. 26
=  
2
x3
3
x
23
?
 – 6x+9log(x+2)+c
It is sometime possible by a change of independent variable to transform a function into another
which can be readily integrated.
We can show the following rules.
To put z = f (x) and also adjust dz = f
 ? 
(x) dx
Example: 
, dx ) x( ' h } ) x( h { F
?
 take e
z
 = h(x) and to adjust dz = h
?
 (x) dx
then integrate 
?
dzz F )( using normal rule.
Example:  
?
? dx ) 3 (2x
7
We put (2x + 3) = t   ?   so 2 dx = dt    or  dx = dt / 2
        Therefore 
? ?
? ?
?
?
? ? ? ? ?      c
16
3x 2
16
t
      
2x8
t
    dt t ½  dx ) 3 (2x
8 8 8
7 7
This method is known as Method of Substitution
Example:  
??
?
?
dx
1x
x
3
2
3
We put  (x
2
 +1) = t
so  2x dx = dt    or  x dx = dt / 2
=   
?
dx
t
x .x
3
2
=   
?
?
dt
t
1 t
2
1
3
=   
? ? 3 2
t
dt
2
1
–
t
dt
2
1
=   
?? ) 13 (–
t
2
1
–
12 –
t
2
1
13 – 12 –
?
?
?
?
? ?
= – 
2
1
  
t
1
 + 
4
1
 
2
t
1
=  
t
1
2
1
t
1
4
1
2
?
© The Institute of Chartered Accountants of India
8. 27 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
    =   
??
??
2
1
1
2
2
11 1
. – . + c
4 2
x + 1
x ?
IMPORTANT STANDARD FORMULAE
a)  
ax
a– x
log
a2
1
a –x
dx
22
?
?
?
 + c
b)  
? 22
a+x dx 1
= log
2a a – x a –x
 + c
c)  
?
?
22
22
dx
=log x + x + a
xa
 + c
d)  ? ?
?
? ? ?
?
22
22
a xx log
ax
dx
 + c
e)  
) x ( f e dx )} x( ’ f ) x( f { e
x x
? ?
?
 + c
f)   
? ?
?
? ? ? ? ? ?
22
2
22 22
a xx log
2
a
ax
2
x
dx ax
 + c
g)  
?
? ? ) – a x   x ( log  
2
a
-     a - x   
2
x
  dx  a - x
22
2
22 22
+ c
h)  
?
    dx  
f(x)
(x) f'
=log f(x) + c
Examples: ( a) 
? ?
?
?
?
22 x2
x
2 z
dz
dx
4 e
e
  where  z=e
 x
  dz = e
 x
 dx
?
?
?
?
?
?
?
?
?
?
?
2e
2e
log
4
1
x
x
  +c
(b)
dx
) 1 – x – x( ) 1 x x(
1 xx
dx
1 xx
1
2 2
2
2
? ?
? ?
? ?
?
? ?
= 
?
? ? dx ) 1 xx (
2
=
c ) 1 x x log(
2
1
1x
2
x
2
x
2 2
2
? ? ? ?? ?
(c)
? ?
? ?
? ? ? ?
3 x 23 x
x ) x ( f where , dx ) x ( ' f ) x ( f e dx ) x 3 x ( e
[by (e) above)]  = e
x
x
3
+c
? ? ? ?
? ? dx ] dx v
dx
) u( d
[ dxv u dxv u
© The Institute of Chartered Accountants of India