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Until now we have been finding integrals of continuous

functions over closed intervals.

Sometimes we can find integrals for functions wherethe function or the limits are infinite. These are calledimproper integrals.

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Example 1:

1

0

1 1

x dxx

The function isundefined atx = 1 .

0

1

2

3

4

1

Sincex = 1 is an asymptote, thefunction has no maximum.

Can we findthe area under

an infinitelyhigh curve?

We could define this integral as:

01

1lim

1

b

b

xdx

x

(left hand limit)We must approach the limit frominside the interval.

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01

1lim

1

b

b

xdx

x

1

1

1

1

x

x

xdx

x

Rationalize the numerator.

2

1+x

1

dx

x

2 2

1 x

1 1

dx dx

x x

21u x

2du x dx 1

2

du x dx 1

1 21

sin2

x u du

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2 2

1 x

1 1

dx dx

x x

2

1u x

2du x dx 1

2

du x dx 1

1 21

sin

2

x u du

1

1 2sin x u

1 2

1 0

lim sin 1

b

b

x x

1 2 1

1

lim sin 1 sin 0 1b

b b

1

2

2

0 0

This integral convergesbecause it approaches asolution.

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Example 2:

-1

0

1

2

3

4

-1 1

1

0

dx

x

1

0

lim lnb

b

x

0lim ln1 lnb

b

0

1

lim lnb b

This integral diverges.

(right hand limit)

We approach the limit from insidethe interval.

1

0

1limbb

dxx

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Example 3:

3

20

31

dx

x The functionapproaches

when .

1x

0

1

2

3

4

1 2 3

2

3

3

0

1x dx

2 23

3 3

01 1

lim 1 lim 1b

cb c

x dx x dx

31 1

3 3

1 10

lim 3 1 lim 3 1

b

b cc

x x

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2 2

3

3 3

01 1

lim 1 lim 1b

cb c

x dx x dx

3

1 1

3 3

1 10

lim 3 1 lim 3 1

b

b cc

x x

11 1 1

33 3 3

1 1

lim 3 1 3 1 lim 3 2 3 1b c

b c

00

33 3 2

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0

1

2

3

4

1 2 3

Example 3 without limits:

3

20

31

dx

x The functionapproaches

when .

1x

2

3

3

0

1x dx

123132313 33330

3 x

So yes, you can evaluate this without limits.

Note that the limit method will alwaysalways work if you

have a singularity. The regular method may not, ormay give you a wrong answer.

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Example 4:

1P

dx

x

0P

1

P

x dx

1

limb

P

b

x dx

1

1

1lim

1

b

P

b

xP

1 11

lim

1 1

P P

b

b

P P

What happens here?

If , then the ratio involvinggets bigger and bigger as ,

therefore the integral diverges.

1P 1Pb

b

If then b has a negativeexponent and ,therefore the integral converges.

1P 1

0P

b