improper integrals 8 8

8.8Improper Integrals

Mark Woodard

Furman U

Fall 2010

Mark Woodard (Furman U) 8.8Improper Integrals Fall 2010 1 / 10

Outline

1 An overview

2 Type I integrals: unbounded domains

3 Type II integrals: unbounded integrands

An overview

What makes an integral improper?

Recall that the definite integral ba f (x)dx is only defined for a

bounded function f on a bounded domain [a, b]. Thus the followingintegrals are improper: 0

exdx is improper because the domain, [0,), is unbounded. 10

x1/2dx is improper because the integrand has a vertical

asymptote at 0, which is in the domain of integration. 0

1

x4/3 + x2/3dx is improper because the integrand is unbounded

and the domain is unbounded.

An overview

bounded function f on a bounded domain [a, b]. Thus the followingintegrals are improper:

0

1

An overview

exdx is improper because the domain, [0,), is unbounded.

10

1

An overview

asymptote at 0, which is in the domain of integration.

0

1

An overview

1

Type I integrals: unbounded domains

Type I: unbounded domains

When the domain of integration is unbounded, we must solve theproblem by limits:

af (x)dx = lim

b

ba

f (x)dx ,

whenever this limit exists.

When this limit exists, we say that the improper integrala f (x)dx

converges. When this limit does not exist, we say that the integraldiverges.

Likewise

b

f (x)dx = lima

ba

f (x)dx .

af (x)dx = lim

b

ba

f (x)dx ,

Likewise

b

f (x)dx = lima

ba

f (x)dx .

af (x)dx = lim

b

ba

f (x)dx ,

Likewise

b

f (x)dx = lima

ba

f (x)dx .

af (x)dx = lim

b

ba

f (x)dx ,

Likewise

b

f (x)dx = lima

ba

f (x)dx .

Problem

Evaluate

1

x2dx

Solution

Let b > 1 and observe that b1

x2dx = x1b1

= 1 1b.

Since limb(1 b1) = 1, we say

1x2dx = 1.

Problem

Evaluate

1

x2dx

Solution

x2dx = x1b1

= 1 1b.

1x2dx = 1.

Problem

Evaluate

1

x2dx

Solution

x2dx = x1b1

= 1 1b.

1x2dx = 1.

Problem

Evaluate

1

x2dx

Solution

x2dx = x1b1

= 1 1b.

1x2dx = 1.

Problem

Evaluate

1

x1dx

Solution

Let b > 1. Then b1

x1dx = ln(x)b1

= ln(b) ln(1) = ln(b).

Since limb ln(b) = +, we say that the integral diverges.However we will sometimes write

1x1dx = +

to indicate that the integral diverges in this particular way.

Problem

Evaluate

1

x1dx

Solution

Let b > 1. Then b1

x1dx = ln(x)b1

= ln(b) ln(1) = ln(b).

1x1dx = +

Problem

Evaluate

1

x1dx

Solution

Let b > 1. Then b1

x1dx = ln(x)b1

= ln(b) ln(1) = ln(b).

1x1dx = +

Problem

Evaluate

1

x1dx

Solution

Let b > 1. Then b1

x1dx = ln(x)b1

= ln(b) ln(1) = ln(b).

1x1dx = +

Problem

Evaluate the following integrals:

0

exdx 0

x2exdx

1

1 + x2dx How should we define this?

0

1

x2 + 3x + 2dx

Problem

Evaluate the following integrals: 0

exdx

0

x2exdx

1

0

1

x2 + 3x + 2dx

Problem

exdx 0

x2exdx

1

0

1

x2 + 3x + 2dx

Problem

exdx 0

x2exdx

1

0

1

x2 + 3x + 2dx

Problem

exdx 0

x2exdx

1

0

1

x2 + 3x + 2dx

Type II integrals: unbounded integrands

If f is unbounded as x a+, then we define ba

f (x)dx = limca+

bc

f (x)dx ,

provided the limit exists.

If f is unbounded as x b, then we define ba

f (x)dx = limcb

ca

f (x)dx ,

f (x)dx = limca+

bc

f (x)dx ,

f (x)dx = limcb

ca

f (x)dx ,

f (x)dx = limca+

bc

f (x)dx ,

f (x)dx = limcb

ca

f (x)dx ,

Problem

Evaluate

10

x1/2dx.

Solution

Let 0 < c < 1 and observe that 1c

x1/2dx = 2x1/21c

= 2 2c .

Since limc0+(2 2c) = 2.

Thus

10

x1/2dx = 2.

Problem

Evaluate

10

x1/2dx.

Solution

x1/2dx = 2x1/21c

= 2 2c .

Thus

10

x1/2dx = 2.

Problem

Evaluate

10

x1/2dx.

Solution

x1/2dx = 2x1/21c

= 2 2c .

Thus

10

x1/2dx = 2.

Problem

Evaluate

10

x1/2dx.

Solution

x1/2dx = 2x1/21c

= 2 2c .

Thus

10

x1/2dx = 2.

Problem

Evaluate

10

x1/2dx.

Solution

x1/2dx = 2x1/21c

= 2 2c .

Thus

10

x1/2dx = 2.

Problem

18

x2/3dx How shall we define this? 10

x21 x2 dx

Problem

Here is an integral of Type I and Type II. Evaluate

0

1

x4/3 + x2/3dx

Problem 18

x2/3dx How shall we define this?

10

x21 x2 dx

Problem

0

1

x4/3 + x2/3dx

Problem 18

x21 x2 dx

Problem

0

1

x4/3 + x2/3dx

Problem 18

x21 x2 dx

Problem

0

1

x4/3 + x2/3dx

An overviewType I integrals: unbounded domainsType II integrals: unbounded integrands

improper integrals 8 8

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