8.8Improper Integrals

Extra Practice

Problems

Until now we have been finding integrals of continuous functions over closed intervals.

Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals.

Example 1:

1

0

1 1x dxx

The function is

undefined at x = 1 .

Since x = 1 is an asymptote, the function has no maximum.

Can we find the area under an infinitely high curve?

We could define this integral as:

01

1lim 1

b

b

x dxx

(left hand limit)

We must approach the limit from inside the interval.

01

1lim 1

b

b

x dxx

1 1

11

xx

x dxx

Rationalize the numerator.

2

1+x 1

dxx

2 2

1 x 1 1

dx dxx x

21u x

2 du x dx1 2du x dx

11 21sin

2x u du

2 2

1 x 1 1

dx dxx x

21u x

2 du x dx1 2du x dx

11 21sin

2x u du

1

1 2sin x u

1 2

1 0lim sin 1

b

bx x

1 2 1

1lim sin 1 sin 0 1b

b b

1

2

2

0 0

This integral converges because it approaches a solution.

Example 2:

1

0

dxx

1

0lim ln

bbx

0lim ln1 lnb

b

0

1lim lnb b

This integral diverges.

(right hand limit)

We approach the limit from inside the interval.

1

0

1lim bb

dxx

Example 3:

3

2031

dx

x

The function approacheswhen .

1x

233

01 x dx

2 233 3

01 1lim 1 lim 1

b

cb cx dx x dx

31 1

3 31 1

0

lim 3 1 lim 3 1b

b cc

x x

2 233 3

01 1lim 1 lim 1

b

cb cx dx x dx

31 1

3 31 1

0

lim 3 1 lim 3 1b

b cc

x x

11 1 133 3 3

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

b c

0 0

33 3 2

Example 4:

1 P

dxx

0P

1 Px dx

1lim

b P

bx dx

1

1

1lim 1

bP

bx

P

1 11lim 1 1

P P

b

bP P

What happens here?

If then gets bigger and bigger as , therefore the integral diverges.

1P 1Pb

b

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

1P 1 0Pb

(P is a constant.)