Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - Review - True-False Quiz: 11



Work Step by Step

Consider $f(x)=\frac{1}{x}$ This function is continuous $[1,\infty)$ and is also decreasing on $[1,\infty)$ (because $f'(x)=\frac{1}{x^{2}}$ , which is always negative for any $x\ne 0$ ). Also $\lim\limits_{n \to \infty}f(x)=0$ However, $\int^{\infty}_{1}f(x)dx =\lim\limits_{x' \to \infty}\int^{x'}_{1}f(x)dx$ $=\lim\limits_{x' \to \infty}\int^{x'}_{1}\frac{1}{x}dx$ $=\lim\limits_{x' \to \infty} [lnx]^{x'}_{1}$ $=\lim\limits_{x' \to \infty} [lnx'-ln1]$ $=\lim\limits_{x' \to \infty} lnx'$ $=\infty $ Therefore, this integral is not convergent. Hence, the given statement is false.
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