This ppt includes all the Topics of this chapter. IMPROPER INTEGRALS AND APPLICATION OF INTEGRATION ppt of Calculus Sem 1. Intergration Sector!

1. IMPROPER INTEGRALS AND
APPLICATION OF
INTEGRATION

2.  DHRUV MISTRY – 160280106055
 SUJAY SHETH – 160280106111
 DHRUV SHAH – 160280106106
 CHIIRAG MAKWANA – 160280106051
 MIHIR PRAJAPATI – 160280106093
 RUTVIK PATEL – 160280106080
 NIYATI SHAH – 160280106109
 SMIT DOSHI – 160280106025
 KEVIN SUTARIA – 160280106115
 ALPESH PRAJAPATI – 160280106088
 ZEEL PATEL – 160280106086
 NIYATI SHAH –
The Team :

3. Improper Integral
TYPE-I:
Infinite Limits of Integration


1 2
1
dx
x 
1
1 2
1
dx
x
TYPE-II:
Discontinuous Integrand
Integrands with Vertical
Asymptotes
Example
Example

4. TYPE-I Infinite Limits



dxxf )( 



a
a
dxxfdxxf )()(
  
 

b
a
b
dxxfdxxf
a
)(lim)(
  



b
aa
dxxfdxxf
b
)(lim)(

5. CONVERGENT AND DIVERGENT 1st
kind
• The improper integrals and
are called:
– Convergent if the corresponding limit exists.
– Divergent if the limit does not exist.
( )
a
f x dx

 ( )
b
f x dx


6. Example 1
• Determine whether the integral is convergent
or divergent.
• Solution:
• According to part (a) of Definition 1, we have
• The limit does not exist as a finite number and so the
• improper integral is divergent.

7. Type-II : Discontinuous Integrands

8. • Find
• Solution:
• We note first that the given integral is improper because
• has the vertical asymptote x = 2.
• Since the infinite discontinuity occurs at the left endpoint
of
• [2, 5], we use part (b) of Definition 3:

9. IMPROPER INTEGRALS

10. IMPROPER INTEGRALS

11. Improper Integral Of Third Kind
It is a definite integral in which one or both limits
of integration are infinite, and the integrand
becomes infinite at one or more points within or
at the end points of the interval of integration.
Thus, it is combination of First and Second Kind.

12. Integral over unbounded
intervals :
•We have definite integrals of the form
and as improper integrals.
•If f is continuous on [a,∞],then is a
proper integral for any b>a.


a
dxxf )(
)()( xdxf
b
a


b
a
dxxf )(
)()()( xdxfdxxf
b
aa
b



  

13. If it is continuous on and if
and both converge , we say that
),(  
a
dxxf )(


a
dxxf )(
  

 


a
a
dxxfdxxfdxxf )()()(
If either or both the integrals on RHS
of this equation diverge, the integral of
f from

14. Application Of
Definite Integrals

15. VOLUME
Area
Between
Curves
By Slicing
Disk
Method
Washer
Method
By
Cylindrical
Shells

16. Areas Between Curves
y = f(x)
y = g(x)
 The area A of the region bounded
by the curves y=f(x), y=g(x), and
the lines x=a, x=b, where f and g
are continuous and f(x) ≥ g(x) for
all x in [a,b], is
 
b
a
dxxgxfA )]()([

17. General Strategy for Area Between Curves:
1
Decide on vertical or horizontal strips. (Pick whichever is
easier to write formulas for the length of the strip, and/or
whichever will let you integrate fewer times.)
Sketch the curves.
2
3 Write an expression for the area of the strip.
(If the width is dx, the length must be in terms of x.
If the width is dy, the length must be in terms of y.
4 Find the limits of integration. (If using dx, the limits are x
values; if using dy, the limits are y values.)
5 Integrate to find area.

18. Q). Consider the area bounded by the graph of the function
f(x) = x – x2 and x-axis:
The volume of solid is:
30/
)5/04/03/0()5/14/23/1(
)5/23/(
)2(
1
0
533
43
1
0
2







 
xxx
dxxxx
1
 
1
0
22
)( dxxx
© iTutor. 2000-2013. All Rights Reserved

19. Solids of Revolution
• A solid generated by revolving a plane area
about a line in the plane is called a Solid of
Revolution.

20. Volume of Solid Of Revolution
can be in :
Cartesian Form
Parametric Form
Polar Form

21. Cartesian Form
dxyV
b
a
2
  dxxV
d
c
2
 

22. Parametric Form
dt
dt
dx
yV
t
t
2
2
1
 
dt
dt
dx
xV
t
t
2
2
1
 

23. Polar Form



drV sin
3
2 3
2
1
 



drV cos
3
2 3
2
1
 

24. Volume by slicing

b
a
dxxAV )(

25. The volume of a solid of known integrable cross-section
area A(x) from x = a and x = b is

b
a
dxxAV )(
 General Strategy :
1. Sketch the Solid & a typical Cross-Section.
2. Find the Formula for that Cross –Section area A(x).
3. Find Limits of Integration.
4. Integrate A(x) using Fundamental Theorem.

26. Q). A pyramid 3 m high has a square base that is 3 m on a
side. The cross-section of the pyramid perpendicular to the
altitude x m down from the vertex is a square x m on a side.
Find the Volume of the pyramid.
•A formula for A(x). The cross-
section at x is a square x meters
on a side, so its area is 2
)( xxA 

27. Volume by Disk Method
To find the volume of a solid like the one
shown in Figure, we need only
observe that the cross-sectional area A(x) is
the area of a disk of radius R(x), the distance
of the planar region’s boundary from the axis
of revolution. The area is then
So the definition of volume gives :

28. To see how to use the volume of a disk to find the
volume of a general solid of revolution, consider a solid
of revolution formed by revolving the plane region in
Figure about the indicated axis.
Figure
Disk method

29. Q). Volume of a Sphere
The circle
is rotated about the x-axis to generate
a sphere. Find its volume.
We imagine the sphere cut
into thin slices by planes
perpendicular to the x-axis.
The cross-sectional area at a
typical point x between x=a to
x=-a
222
ayx 

30. Volume by Washer Method
• The disk method can be extended to cover solids of revolution
with holes by replacing the representative disk with a
representative washer.
• The washer is formed by revolving a rectangle about an axis ,
as shown in Figure.
• If r and R are the inner and outer radii
of the washer and w is the width of the
washer, the volume is given by
• Volume of washer = π(R2 – r2)w.
Axis of revolution Solid of revolution

31. If the region we revolve to generate a
solid does not border on or cross the axis
of revolution, the solid has a hole in it
(Figure ). The cross-sections perpendicular
to the axis of revolution are washers
instead of disks.
The dimensions of a typical washer are
•Outer radius : R(x)
•Inner radius : r(x)

32. 0
1
2
3
4
1 2
Q). The region bounded by
and is revolved
about the y-axis.
Find the volume.
2
y x 2y x
The “disk” now has a hole in it,
making it a “washer”.
If we use a horizontal slice:
The Volume of the washer is:  2 2
thicknessR r  
 2 2
R r dy 
outer
radius
inner
radius
2y x
2
y
x
2
y x
y x
2
y x
2y x
 
2
24
0 2
y
V y dy
  
      

4
2
0
1
4
V y y dy
 
  
 

4
2
0
1
4
V y y dy 
4
2 3
0
1 1
2 12
y y
 
   
16
8
3

 
   
8
3




33. Disks v/s Washers
 
d
c
dyyvywV ))]([)](([ 22

d
c
dyyuV 2
)]([

34. VOLUME BY
CYLINDRICAL SHELL

36. 36
General Strategies :
• Sketch the graph over the limits of integration
• Draw a typical shell parallel to the axis of revolution
• Determine radius, height, thickness of shell
• Volume of typical shell
• Use integration formula
2 radius height thickness   
2
b
a
Volume radius height thickness   

37. VOLUMES BY CYLINDRICAL SHELLS
Cylinder
h
r2
rh2Areasurface 
Shell
rrh 2volumeshell
r
Cylinder
h
r2
)(),( 122
1
12 rrrrrr 

38. Q). Find the volume of the solid obtained by rotating about
the y-axis the region between y = x and y = x2.
• We see that the shell has radius x, circumference 2πx, and
height x - x2.
• Thus, the volume of the solid is:
  
 
1
2
0
1
2 3
0
13 4
0
2
2
2
3 4 6
V x x x dx
x x dx
x x




 
 
 
   
 



39. Method Axis of rotation Formula
Disks and
Washers
The x-axis
The y-axis
Cylindrical
Shells
The x-axis
The y-axis