IMPROPER INTEGRALS AND

APPLICATION OF INTEGRATION

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Improper Integral

TYPE-I:Infinite Limits of Integration

1 2 1 dxx

1

1 2 1 dxx

TYPE-II:Discontinuous IntegrandIntegrands with Vertical

AsymptotesExample

Example

TYPE-I Infinite Limits

dxxf )(

a

a dxxf dxxf )()(

b

a

b

dxxfdxxfa

)(lim)(

b

aa

dxxfdxxfb

)(lim)(

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

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.

Type-II : Discontinuous Integrands

• 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:

IMPROPER INTEGRALS

IMPROPER INTEGRALS

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.

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 )(

)()( xdxfb

a

b

a

dxxf )(

)()()( xdxfdxxfb

aa

b

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

Application Of Definite Integrals

VOLUME

Area Between Curves

By SlicingDisk Method

Washer Method

By Cylindrical Shells

Areas Between Curvesy = 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

adxxgxfA )]()([

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.

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

431

0

2

xxx

dxxxx

1

1

0

22 )( dxxx

© iTutor. 2000-2013. All Rights Reserved

Solids of Revolution• A solid generated by revolving a plane area

about a line in the plane is called a Solid of Revolution.

Volume of Solid Of Revolution can be in :

Cartesian Form

Parametric Form

Polar Form

Cartesian Form

dxyVb

a

2 dxxVd

c

2

Parametric Form

dtdtdxyV

t

t

22

1

dtdtdxxV

t

t

22

1

Polar Form

drV sin32 3

2

1

drV cos32 3

2

1

Volume by slicing

b

a

dxxAV )(

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.

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

Volume by Disk MethodTo find the volume of a solid like the one shown in Figure, we need onlyobserve that the cross-sectional area A(x) is the area of a disk of radius R(x), the distanceof the planar region’s boundary from the axis of revolution. The area is then

So the definition of volume gives :

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

Q). Volume of a SphereThe circleis 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

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

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)

Q). The region bounded by and is revolved about the y-axis.Find the volume.

2y 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 2R r dy

outerradius

innerradius

2y x

2y x

2y x

y x

2y x

2y x

2

24

0 2yV y dy

4 2

0

14

V y y dy

4 2

0

1 4

V y y dy 4

2 3

0

1 12 12

y y

1683

83

Disks v/s Washers

d

c

dyyvywV ))]([)](([ 22d

c

dyyuV 2)]([

VOLUME BY CYLINDRICAL SHELL

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

2b

a

Volume radius height thickness

VOLUMES BY CYLINDRICAL SHELLS

Cylinder

h

r2rh2Area surface

Shell

rrh 2 volumeshell

r

Cylinder

h

r2

)(),( 1221

12 rrrrrr

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

23 4 6

V x x x dx

x x dx

x x

Method Axis of rotation Formula

Disks and

Washers

The x-axis

The y-axis

CylindricalShells

The x-axis

The y-axis

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YOU!