Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises: 5

Answer

$$\int_1^\infty x^{-2} dx=1$$

Work Step by Step

To evaluate the improper integral recall that $$\int_a^\infty f(x) dx =\lim\limits_{b \to \infty} \int_a^bf(x)dx$$ So for our problem we have: $$\int_1^\infty x^{-2} dx =\lim\limits_{b \to \infty} \int_1^bx^{-2}dx$$ Compute the integral: $$\lim\limits_{b \to \infty} \int_1^bx^{-2}dx=\lim\limits_{b \to \infty}\frac{x^{-1}}{-1}\bigg |_1^b=\lim\limits_{b \to \infty}\left(-\frac{1}{b}+1 \right)=0+1=1$$
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