Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 461: 78


The integral converges and its value is $\frac{5}{7}\frac{1}{9^{7/5}}$.

Work Step by Step

We have $$ \int_{9}^{\infty} \frac{d x}{x^{12/5}}= \int_{9}^{\infty}x^{-12/5}dx\\ =-\frac{1}{7/5}x^{-7/5}|_9^{\infty}=-\frac{5}{7}(\frac{1}{\infty}-\frac{1}{9^{7/5}})=\frac{5}{7}\frac{1}{9^{7/5}} $$ Hence, the integral converges and its value is $\frac{5}{7}\frac{1}{9^{7/5}}$.
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