8.4 Improper Integrals AP Calculus BC. 8.4 Improper Integrals One of the great characteristics of...

8.4 Improper Integrals

AP Calculus BCAP Calculus BC

8.4 Improper Integrals

One of the great characteristics of mathematics is that mathematicians are constantly finding ways to get around the rules, or to bend the rules, or just plain ignore them.

Improper integrals is a technique to use when an interval is not finite, and when an integrand is not continuous.

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.

First, here are the definitions of Improper Integrals with Infinite Solutions:

1) If is continuous on , , then

limb

a ab

f x a

f x dx f x dx

2) If is continuous on , , then

limb b

aa

f x b

f x dx f x dx

If the Limit is finite, then the Improper Integral converges.

If the Limit fails, then it diverges

3) If is continuous on , , thenc

c

f x

f x dx f x dx f x dx

If both Improper Integrals

converge, then so does .f x dx

Express the improper integral as a limit of definite integrals and evaluate the integral.

22

2

1

xdx

x

First, rewrite and split up the integral.

22

2

1

xdx

x

22

0 2

1

xdx

x

220

2

1

xdx

x

0

22

2lim

1aa

xdx

x

20 2

2lim

1

b

b

xdx

x

2

0

1lim

1

b

b x

2 20

1 1lim

1 1b b

1

The antiderivative of 20 2

2

1

b xdx

x

= 0 as b → ∞

Express the improper integral as a limit of definite integrals and evaluate the integral.

22

2

1

xdx

x

Now, by analyzing the integrand, the denominator is always positive for all x. [i.e., (x2 + 1)2 is always positive.]

So, 22

2

1

x

x is positive when x > 0 and negative when x < 0.

Therefore,

0

22

21

1

xdx

x

because x < 0

and, 22

2

1

xdx

x

= −1 + 1 = 0

2

0

1lim

1

b

b x

2 2

1 1lim

1 0 1b b

1

Evaluate the following integral or state that it diverges.1

xxe dx

Use Integration by Parts u dv uv v du u xdu dx

xdv e dxxv e

1

xxe dx 1

bxx e 1

b xe dx 1

bx xx e e

1

1bxe x 11 2be b e

1 2lim

bb

b

e e

= 0

2

e

Improper Integrals with Infinite Discontinuities

Integrals of functions that become infinite at a point within the interval of integration are improper integrals.

1) If is continuous on a, b , then

limb b

a cc a

f x

f x dx f x dx

2) If is continuous on , , then

limb c

a ac b

f x a b

f x dx f x dx

3) If is continuous on , c, b , thenb c b

a a c

f x a c

f x dx f x dx f x dx

Another type of improper integral arises when the integrand has a vertical asymptote. And that’s also a point of infinite discontinuity either at a limit of integration or at some other point between the limits of integration.

Finite Limit Converges

Limit Fails Diverges

Both Limits Finite Converges

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+

1

xdx

x

2 2

1

1 1

xdx 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

2 233 3

01 1lim 1 lim 1

c

cc cx dx x dx

31 1

3 31 1

0

lim 3 1 lim 3 1c

c cc

x x

3

2031

dx

x

1 3

2 20 13 31 1

dx dx

x x

2 233 3

01 1lim 1 lim 1

c

cc cx dx x dx

31 1

3 31 1

0

lim 3 1 lim 3 1c

c cc

x x

11 1 133 3 3

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

c 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.)

Day 1

Theorem 6 Comparison Test

Let and be continuous on , with 0 for all . Thenf g a f x g x x a

1. converges if converges.a af x dx g x dx

2. diverges if diverges.a ag x dx f x dx

1

xe dx

1lim

b x

be dx

1lim

bx

be

1lim b

be e

1 1lim

bb e e

0 1

e

Converges

Does converge?2

1

xe dx

Compare:

to for positive values of x.2

1xe

1xe

For2

2

1 11,

ex x

xxx e e

e

2

1xe

1xe

For2

2

1 11,

ex x

xxx e e

e

Since is always below , we say that it is

“bounded above” by .

2

1xe

1xe1

xe

Since converges to a finite number, must also converge!1xe

2

1xe

Example 7:

2

21

sin xdx

x

The maximum value of so:sin 1x

2

2 2

sin 10 on 1,

x

x x on

Since converges, converges.2

1

x

2

2

sin x

x

Example 7:

21

1

0.1dx

x

for positive values of x, so:2 0.1x x

Since diverges, diverges.1

x 2

1

0.1x

2

1 1

0.1 xx

on 1,

If functions grow at the same rate, then either they both converge or both diverge.

Does converge?21 1

dx

x

As the “1” in the denominator becomes

insignificant, so we compare to .

x

2

1

x

2

2

1

lim1

1

x

x

x

2

2

1limx

x

x

2lim

2x

x

x 1 Since converges,

converges.

2

1

x2

1

1 x

21 1

dx

x

Of course

21lim

1

b

b

dx

x

1

1lim tan

b

bx

1 1lim tan tan 1b

b

2 4

4

tany x

As , 2

y x

2

2

4

4

21 1

dx

x

21lim

1

b

b

dx

x

1

1lim tan

b

bx

1 1lim tan tan 1b

b

2 4

4

Of course 21

1 dx

x

2

1lim

b

bx dx

1

1lim

b

bx

1 1lim

1b b

1

0

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