7.8 Integration Techniques

Improper Integrals

Question #5 (Medium): Convergent Improper Integral With Infinite Discontinuity

Strategy

When dealing with type improper integrals, the interval usually does not contain infinity but is over

numeric values. It is categorized as improper integral of type because asymptotic value is caught in the

interval. To apply the limit, the asymptotic value needs to be determined first. The value of that makes

the denominator is the value the limit should approach. Once the limit is set, split the interval with

respect to the asymptote, or if one end of the interval is the asymptote, replace that value with variable

. Then disregard the limit and evaluate the integral. Add the anti-derivative to the limit, then plug in the

approaching value to see if it diverges or converges. Sometimes directly plugging in the value cannot be

done. Then l’Hospital’s rule can be applied.

Sample Question

Determine if the integral is convergent or divergent. If convergent, evaluate the integral.

Solution

does not exist. It is the asymptote. Therefore the beginning of the interval is associated with the

limit. Since approaches the asymptote from the right, limit is set to approach from the right side

with the superscript . Then:

. After setting the limit, focus on solving

the integral alone. The two distinctively different functions can be solved by integration by parts.

Though simplifies with derivative, there is no other way to deal with . So let ,

,

and , and

. Integration by parts does not require the interval to be changed. So:

Then add the limit and solve:

. Now problem arises because expression is

still part of the anti-derivative. Take the whole expression , and apply l’Hospital’s rule:

Therefore,

Thus, the improper integral is convergent and

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###### Document Summary

Question #5 (medium): convergent improper integral with infinite discontinuity. When dealing with type improper integrals, the interval usually does not contain infinity but is over numeric values. It is categorized as improper integral of type because asymptotic value is caught in the interval. To apply the limit, the asymptotic value needs to be determined first. The value of that makes the denominator is the value the limit should approach. Once the limit is set, split the interval with respect to the asymptote, or if one end of the interval is the asymptote, replace that value with variable. Then disregard the limit and evaluate the integral. Add the anti-derivative to the limit, then plug in the approaching value to see if it diverges or converges. Sometimes directly plugging in the value cannot be done. Determine if the integral is convergent or divergent. Therefore the beginning of the interval is associated with the limit.

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