Math 104: Improper Integrals (With Solutions)

[Pages:20]Math 104: Improper Integrals (With Solutions)

Ryan Blair

University of Pennsylvania

Tuesday March 12, 2013

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 1 / 15

Outline

1 Improper Integrals

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 2 / 15

Improper Integrals

Improper integrals

b

Definite integrals f (x)dx were required to have

a

finite domain of integration [a, b] finite integrand f (x) < ?

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 3 / 15

Improper Integrals

Improper integrals

b

Definite integrals f (x)dx were required to have

a

finite domain of integration [a, b] finite integrand f (x) < ?

Improper integrals 1 Infinite limits of integration 2 Integrals with vertical asymptotes i.e. with infinite discontinuity

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 3 / 15

Improper Integrals

Infinite limits of integration

Definition Improper integrals are said to be

convergent if the limit is finite and that limit is the value of the improper integral. divergent if the limit does not exist.

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 4 / 15

Improper Integrals

Infinite limits of integration

Definition Improper integrals are said to be

convergent if the limit is finite and that limit is the value of the improper integral. divergent if the limit does not exist.

Each integral on the previous page is defined as a limit.

If the limit is finite we say the integral converges, while if the limit is infinite or does not exist, we say the integral diverges.

Convergence is good (means we can do the integral); divergence is bad (means we can't do the integral).

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 4 / 15

Example 1

Improper Integrals

Find (if it even converges)

e-x dx .

0

Solution:

b

b

e-x dx = lim e-x dx = lim - e-x

0

b 0

b

0

= lim -e-b + e0 = 0 + 1= 1.

b

So the integral converges and equals 1.

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 5 / 15

Example 2

Improper Integrals

Find (if it even converges)

-

1 1 + x2

dx .

Solution: By definition,

1

c1

1

- 1 + x 2 dx = - 1 + x 2 dx + c 1 + x 2 dx ,

where we get to pick whatever c we want. Let's pick c = 0.

0 -

1 1 + x2

dx

= lim b-

0

arctan(x) = lim [arctan(0) - arctan(b)] b b-

= 0 - lim arctan(b) =

b-

2

Ryan Blair (U Penn)

Math 104: Improper Integrals

Tuesday March 12, 2013 6 / 15

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