silemus.

32Improper integral of 1/x21/x^2

Evaluate the improper integral ∫1∞1x2 dx\displaystyle \int_1^{\infty} \frac{1}{x^2}\,dx.

Replace infinity with t, integrate, then let t go to infinity. Antiderivative is −1/x, so the integral is (1 − 1/t) → 1 as t→∞. The area is finite and equals 1.

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