CA Foundation Exam  >  CA Foundation Notes  >  ICAI Notes- Integral Calculus

ICAI Notes- Integral Calculus - CA Foundation PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 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
Read More

FAQs on ICAI Notes- Integral Calculus - CA Foundation

1. What is integral calculus?
Ans. Integral calculus is a branch of mathematics that deals with the concept of integration and the calculation of definite and indefinite integrals. It focuses on finding the antiderivative of a function and determining the area under a curve.
2. How is integral calculus relevant to the CA Foundation exam?
Ans. Integral calculus is an essential topic in the CA Foundation exam, specifically in the subject of Mathematics. Questions related to definite and indefinite integrals, integration by substitution, and application of integrals are commonly asked in the exam.
3. What are definite and indefinite integrals?
Ans. Definite integrals are used to calculate the exact area under a curve between two specified points. It gives a specific numerical value. On the other hand, indefinite integrals represent the antiderivative of a function and yield a general solution, as it includes a constant of integration.
4. How can I solve integration problems effectively?
Ans. To solve integration problems effectively, it is crucial to understand the basic integration rules and techniques such as integration by substitution, integration by parts, and partial fractions. Practicing a variety of integration problems and familiarizing yourself with different types of functions will also improve your problem-solving skills.
5. What are some real-life applications of integral calculus?
Ans. Integral calculus has various real-life applications, including finding the area and volume of irregular shapes, calculating work done by a force, determining the center of mass, and solving problems related to growth and decay rates. It is also used in physics, engineering, economics, and other fields to model and analyze continuous processes.
Download as PDF

Top Courses for CA Foundation

Related Searches

Previous Year Questions with Solutions

,

mock tests for examination

,

Objective type Questions

,

Exam

,

Viva Questions

,

ICAI Notes- Integral Calculus - CA Foundation

,

MCQs

,

practice quizzes

,

shortcuts and tricks

,

ppt

,

study material

,

Extra Questions

,

past year papers

,

ICAI Notes- Integral Calculus - CA Foundation

,

Semester Notes

,

video lectures

,

pdf

,

Important questions

,

ICAI Notes- Integral Calculus - CA Foundation

,

Free

,

Summary

,

Sample Paper

;