AP® Calculus BC · Unit 6 · Topic 6.13

Improper Integrals with an Infinite Discontinuity

An integral is improper when the function has a vertical asymptote somewhere in the interval; you handle it by replacing the bad endpoint with a variable and taking a limit, which either settles to a finite number or blows up.

The Fundamental Theorem of Calculus lets you evaluate an integral by plugging endpoints into an antiderivative, but that shortcut assumes the function is continuous everywhere on the interval. If the curve shoots off to infinity at some point c in [a, b], the region under it near that point might still have finite area, or it might not. You cannot tell just by plugging in, because the antiderivative itself is often undefined or discontinuous right at that point. You have to check carefully, using a limit.

The fix is to rewrite the integral as a limit. If the discontinuity sits at the upper limit b, replace b with a variable t, integrate from a to t, and take the limit as t approaches b from the left. If the discontinuity is at the lower limit, do the same from the right. If it sits inside the interval, split the integral into two pieces at that point and take two separate limits, one for each side; the whole thing only converges if both limits exist and are finite. Then evaluate using the antiderivative on the safe, finite side, where it stays continuous, and let the limit variable creep toward the asymptote.

This topic is where integrals meet the limit ideas from earlier units, and it sets up the convergence tests you use on improper integrals with infinite bounds. The same instinct, replace the dangerous value with a letter and take a limit, carries forward into comparing series and integrals for convergence.

Key terms

Improper integral
An integral where the function has a vertical asymptote in the interval, or the interval itself is infinite.
Infinite discontinuity
A point where a function's value grows without bound, creating a vertical asymptote.
Converges
Describes an improper integral whose limit exists and equals a finite number.
Diverges
Describes an improper integral whose limit grows without bound or fails to exist.

Watch it explained

  1. How it worksHow do I split an integral at a vertical asymptote and take two separate limits?
  2. Worked exampleFind the integral from 0 to 1 of 1/√x dx.
  3. Worked exampleFind the integral from 0 to 3 of 1/(x - 1)2 dx and show me what goes wrong.
  4. Common mistakeWhy isn't plugging endpoints into the antiderivative valid across a vertical asymptote?

Check yourself

Your turn1 of 2
Find the integral from 0 to 7 of 1/√x dx.

End of topic.