The integral is evaluated by writing it as

MATHEMATICS · Single correct · AP · Calculus BC · Integration and Accumulation of Change · Integration and Accumulation of Change

Problem

The integral ∫−221x2 dx\displaystyle \int_{-2}^{2} \frac{1}{x^{2}}\,dx is evaluated by writing it as ∫−201x2 dx+∫021x2 dx\displaystyle \int_{-2}^{0} \frac{1}{x^{2}}\,dx+\int_{0}^{2} \frac{1}{x^{2}}\,dx. Which of the following statements best justifies the conclusion that the original integral diverges? (a) Because the integrand is even, the infinite contributions from the left and right of x=0x=0 cancel each other, so the integral actually converges to 00. (b) If either of the two improper limits created in the split fails to exist, then the entire integral fails to exist; one of those limits fails to exist, so the original integral diverges. (c) Evaluating the antiderivative −1/x-1/x directly from x=−2x=-2 to x=2x=2 gives −12−(−1−2)=−1-\tfrac12-(-\tfrac1{-2})=-1, a finite number; therefore the integral converges. (d) Since 1/x21/x^{2} is of the form x−px^{-p} with p=2>1p=2>1, the pp-test guarantees the improper integral converges on (−2,2)(-2,2).

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