integeration.ppt

Introduction
Team Members- There are two members.
• Krishna Singh
• Akshay Singh
Note:- We are here to present the presentation about our 6th
unit, Which name is Integration or Anti-Differentiation.
Class- BCA(Bachelor Of Computer Application) First
Year
1

Integration or Anti-Differentiation
Definition-Integration is a method of adding or
summing up the parts to find the whole. It is a reverse
process of differentiation, where we reduce the
functions into parts. This method is used to find the
summation under a vast scale.
Integration is the calculation of an integral. Integrals in
Math are used to find many useful quantities such as
areas, volumes, displacement, etc.
2

3
INTEGRATION
or
ANTI-DIFFERENTIATION

Now , consider the question :” Given that y is a function of x
and
Clearly ,
( differentiation process )
1.1.THE CONCEPT OF INTEGRAL
We have learnt that
2
x
y  x
dx
dy
2

x
dx
dy
2
 , what is the function ? ‘
2
x
y  is an answer but is it the only answer ?

4
Familiarity with the differentiation process would
indicate that
and in fact

50
,
3
,
15
,
3 2
2
2
2







 x
y
x
y
x
y
x
y
c
x
y 
 2
Thus x
dx
dy
2
 c
x
y 
 2
This process is the reverse process of differentiation
and is called integration .
, where c is can be
any real number are also possible answer
, where c is called an arbitrary constant

5
SYMBOL OF INTEGRATION
We know that
Hence ,
Symbolically , we write

    x
c
x
dx
d
and
x
x
dx
d
2
2 2
2



x
dx
dy
2
 c
x
y 
 2
x
dx
dy
2
  

 c
x
dx
x
y 2
2


6
In general,
then
The expression
 
  )
(
)
( x
f
x
F
dx
d
and
x
F
y
if 

 



 c
x
F
dx
x
f
y
x
f
dx
dy
)
(
)
(
)
(
 

 c
x
F
dx
x
f
y )
(
)
(
Is called an indefinite integral.
 

 c
x
dx
x
y 2
2 Is an indefinite integral

7
When
In general :
When n = 0 ,
  n
n
n
ax
n
n
n
ax
dx
d
dx
x
f
d
n
n
ax
x
F )
1
(
1
1
1
)
(
,
)
1
(
1
)
(
1
1


















n
ax

 






 
1
,
.
1
1 1
n
where
c
ax
n
dx
ax
y
ax
dx
dy n
n
n
 



 c
ax
dx
a
y
a
dx
dy

8
EXAMPLE:
1.
2.The gradient of a curve , at the point ( x,y ) on the curve is
given by
Solution :
Given
   




 dx
x
dx
x
dx
x
dx
x
x
x 2
3
2
3
)
(
c
x
x
x 


 2
3
4
2
1
3
1
4
1
2
4
3
2 x
x 

Given that the curve passes through the point ( 1, 1) ,
find the equation of the curve.
2
4
3
2 x
x
dx
dy


  


 dx
x
x
y )
4
3
2
( 2
  


 dx
x
dx
x
dx 2
4
3
2

9
Since the curve passes through the point ( 1,1 ) , we can
substitusi x = 1 and y = 1 into ( 1 ) to obtain the constant
term c .
The equation of the curve is
)
1
(
......
3
4
2
3
2 3
2
c
x
x
x
y 



c



 )
1
(
3
4
)
1
(
2
3
)
1
(
2
1
6
5


c
c
x
x
x
y 


 3
2
3
4
2
3
2

10
INTEGRATION OF TRIGONOMETRICAL
FUNCTION
If y = sin x then
If y = cos x then
If y = tan x then
If y = sec x then ;
x
dx
dy
cos

x
dx
dy
sin


x
dx
dy 2
sec

x
x
dx
dy
sec
.
tan


11
If y = cot x then
Hence :
ecx
x
dx
dy
cos
.
cot



















c
x
dx
x
ec
x
c
x
dx
x
x
c
x
dx
c
x
dx
x
c
x
dx
cot
cos
.
cot
sec
sec
.
tan
tan
sec
cos
sin
sin
cos
2

12
In each of the above cases , x is measured in radians and c
denotes an arbitrary constant. as in differentiation ,
integration of trigonometrical function is performed only
when the angles involved are measured in radians
In general :
)
cos(
,
)
sin(
1
b
ax
dx
dy
b
ax
a
y
If 



c
b
ax
a
dx
b
ax 



 )
sin(
1
)
cos(
where a , b and c are constant .

13
)
sin(
,
)
cos(
1
b
ax
dx
dy
b
ax
a
y
If 




c
b
ax
a
dx
b
ax 




 )
cos(
1
)
sin(
)
(
sec
,
)
tan(
1 2
b
ax
dx
dy
b
ax
a
y
If 



 


 c
b
ax
a
dx
b
ax )
tan(
1
)
(
sec2
Where a , b and c are constant

14
)
(
cos
,
)
(
cot
1 2
b
ax
ec
dx
dy
b
ax
a
y
If 




 



 c
b
ax
a
dx
b
ax
ec )
cot(
1
)
(
cos 2
)
tan(
).
sec(
,
)
sec(
1
b
ax
b
ax
dx
dy
b
ax
a
y
If 




 


 )
sec(
1
)
tan(
).
sec( b
ax
a
dx
b
ax
b
ax
)
cot(
).
(
cos
,
)
(
1
b
ax
b
ax
ec
dx
dy
b
ax
coesc
a
y
If 





 




 c
b
ax
ec
a
dx
b
ax
b
ax
ec )
(
cos
1
)
cot(
).
(
cos
Where a,b and c are constants

15
Example:
Find The following integrals:
1.
2.
3.
4.
 
 dx
x )
3
5
cos( c
x 
 )
3
5
sin(
5
1
 
 dx
x)
2
3
(
sec2
c
x 

 )
2
3
tan(
2
1
 

 dx
x
x
ec )
3
4
tan(
).
3
4
(
cos c
x
ec 

 )
3
4
(
cos
4
1
 
dx
2
sin 

dx
x
2
2
cos
1
 
 dx
x)
2
cos
1
(
2
1
c
x
x 







 2
sin
2
1
2
1
c
x
x 

 2
sin
4
1
2
1

16
DEFINITE INTEGRAL
Consider f(x) = 3
The area bounded by y = 3 , the line x = a and x = b and the
axis is A = 3 ( b – a )
= 3b – 3a
= F(b) – F(a)
y
x
a b
y = 3
 
 dx
dx
x
f 3
)
(
c
x 
3
3
)
(
,
)
( 

 x
F
where
c
x
F

17
We can write F(b) – F(a) simply as
Further
   b
a
b
a x
or
x
F 3
)
(
  simply
or
dx
x
f
b
a
 )
(  

b
a
b
a
dx
dx
x
f 3
)
(
 b
a
c
x 
 3
 b
a
c
x
F 
 )
(
   
c
a
F
c
b
F 


 )
(
)
(
)
(
)
( a
F
b
F 


18
Similarly , consider f(x) = x + 1
a b
(a,a+1)
1
(b,b+1)
y = x+1
  
 dx
x
dx
x
f )
1
(
)
(
c
x
x



2
2
where
c
x
F ,
)
( 

x
x
x
F 
 2
2
1
)
(

19
The area bounded by y = x + 1 , the lines x = a and x = b ,
and axis is
A = )
)(
1
1
(
2
1
a
b
b
a 


 )
)(
2
(
2
1
a
b
b
a 



)
2
2
(
2
1 2
2
a
a
b
b 


 


















 a
a
b
b
2
2
2
2
)
(
)
( a
F
b
F 

b
a
b
a
x
x
x
dx
x
 









2
)
1
(

20
In general , if
x = a and x = b is given by
 
 c
x
F
dx
x
f )
(
)
(
then the definite integral of f(x) between the limits
 
 


b
a
b
a a
F
b
F
x
F
dx
x
f )
(
)
(
)
(
)
(

21
Example :
Evaluate
1.
2.
 






4
2
2
3 16
dx
x
x  




4
2
2
3
16 dx
x
x
4
2
4
16
4








x
x
)
8
4
(
)
4
64
( 


 56

 
3
0
)
2
sin
3
(

dx
x  3
0
2
cos
3

x
x 


3
2
2
3 

 )
9
4
(
6
1

 

22
SUBSTITUTION ( ALGEBRAIC)
Consider the integral :
We may find this integral by expanding
Suppose that v = 2x + 5 , then
So that
, or we can use the following way.
  dx
x 7
)
5
2
(
 7
5
2 
x
2
2
dv
dx
dx
dv



 
 dx
x 7
)
5
2
(  
2
7 dv
v  
dv
v7
2
1
c
v 
 8
8
.
2
1
c
v 
 8
16
1
c
x 

 8
)
5
2
(
16
1

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