5 Steps to a 5 AP Calculus BC 2019

Definite Integrals 247

Example 1

Evaluate

∫∞

1

1

x

dx.

∫∞

1

1

x

dx=limk→∞

∫k

1

1

x

dx=limk→∞[lnx]k 1 =limk→∞(lnk)=∞so the integral diverges.

Example 2

Evaluate

∫∞

−∞

xe−x^2 dx.

Since both limits of integration are infinite, consider the sum of the improper inte-

grals

∫∞

−∞

xe−x

2

dx=

∫ 0

−∞

xe−x

2

dx+

∫∞

0

xe−x

2

dx. This sum is the sum of the limits

limk→−∞

∫ 0

k

xe−x^2 dx+limc→∞

∫c

0

xe−x^2 dx=klim→−∞

[

−

1

2

e−x^2

] 0

k

+limc→∞

[

−

1

2

e−x^2

]c

0

=

limk→−∞

[

−

1

2

+

1

2

e−k

2

]

+limc→∞

[

−

1

2

e−c

2

+

1

2

]

=−

1

2

+

1

2

= 0 .Since the limit exists, the integral

converges and

∫∞

−∞

xe−x^2 dx=0.

Infinite Discontinuities

If the function has an infinite discontinuity at one of the limits of integration, then
∫b

a

f(x)dx=liml→b−

∫l

a

f(x)dxor

∫b

a

f(x)dx=liml→a+

∫b

l

f(x)dx. If an infinite discontinu-

ity occurs atx=cwithin the interval of integration (a,b), then the integral can be broken
into sections at the discontinuity and the sum of the two improper integrals can be found.
∫b

a

f(x)dx=

∫c

a

f(x)dx+

∫b

c

f(x)dx=limk→c−

∫k

a

f(x)dx+liml→c+

∫b

l

f(x)dx

Example

Evaluate

∫π/ 2

0

√cosx

1 −sinx

dx.

Sincef(x)=
√cosx
1 −sinx

has an infinite discontinuity atx=

π

2

, the integral is improper.

Evaluate

∫π/ 2

0

cosx

√

1 −sinx

dx = klim→π/ 2 −

∫k

0

cosx

√

1 −sinx

dx = klim→π/ 2 −

[

− 2

√

1 −sinx

]k

0

=

klim→π/ 2 −

[

− 2

√

1 −sink+ 2

]

= 2 .Since the limit exists,

∫π/ 2

0

cosx

√

1 −sinx

dx=2.