CK-12-Calculus

http://www.ck12.org Chapter 7. Integration Techniques

∫ 4

0

dx

x− 1

develops an infinite discontinuity atx=1 because the integrand approaches infinity at this point. However, it is
continuous on the two intervals[ 0 , 1 )and( 1 , 4 ].Looking at the integral more carefully, we may split the interval
[ 0 , 4 ]→[ 0 , 1 )∪( 1 , 4 ]and integrate between those two intervals to see if the integral converges.

∫ 4

0

dx

x− 1 =

∫ 1

0

dx

x− 1 +

∫ 4

1

dx

x− 1.

We next evaluate each improper integral. Integrating the first integral on the right hand side,

∫ 1

0

dx

x− 1 =llim→ 1 −

∫l

0

dx

x− 1

=l→lim 1 −[ln|x− 1 |]l 0

=l→lim 1 −[ln|l− 1 |−ln|− 1 |]

=−∞.

The integral diverges because ln( 0 )is undefined, and thus there is no reason to evaluate the second integral. We
conclude that the original integral diverges and has no finite value.
Example 4:
Evaluate∫ 13 √xdx− 1.

Solution:

∫ 3

1

√dx

x− 1 =llim→ 1 +

∫ 3

l

√dx

x− 1

=llim→ 1 +

[

2 √x− 1

] 3

l

=llim→ 1 +

[

2 √ 2 − 2 √l− 1

]

= 2

√

2.

So the integral converges to 2

√

2.

Example 5:
In Chapter 5 you learned to find the volume of a solid by revolving a curve. Let the curve bey=xe−x, 0 ≤x≤∞
and revolving about thex−axis. What is the volume of revolution?
Solution: