8.4 day one

Improper Integrals

Greg Kelly, Hanford High School, Richland, Washington

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

1

xdx

x

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

xdx

x

(left hand limit)

We must approach the limit from inside the interval.

01

1lim

1

b

b

xdx

x

1

1

1

1

x

x

xdx

x

Rationalize the numerator.

2

1+x

1dx

x

2 2

1 x

1 1dx dx

x x

21u x

2 du x dx1

2du x dx

11 2

1sin

2x u du

2 2

1 x

1 1dx dx

x x

21u x

2 du x dx1

2du x dx

11 2

1sin

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

dx

x

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

bbdx

x

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

dx

x

0P

1 Px dx

1lim

b P

bx dx

1

1

1lim

1

bP

bx

P

1 11lim

1 1

P P

b

b

P 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.)