Does ∫_1^∞ 1/x^2 dx converge, and if so, what is its value?

• A. Diverges
• B. Converges; value = 1
• C. Converges; value = 0
• D. Converges; value = ∞

Answer: (B)

This question is about improper integrals and whether a tail integral converges, and if so, what its value is. For the integral from 1 to infinity of 1/x^2, treat it as an improper integral by taking a limit: compute ∫ from 1 to t of 1/x^2 dx and then let t grow without bound.

The antiderivative of 1/x^2 is -1/x. So ∫ from 1 to t of 1/x^2 dx = [-1/x] from 1 to t = (-1/t) - (-1) = 1 - 1/t. Now take the limit as t → ∞: 1 - 0 = 1. Since this limit exists and is finite, the integral converges, and its value is 1.