Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Review - Concept Check - Page 552: 6

Answer

The statement is True: The given improper integral $$ \int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x =\frac{1}{\sqrt {2}-1} $$ is convergent.

Work Step by Step

The given integral $$ \int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x $$ is an improper integral, hence $$ \begin{split} \int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x & = \lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{x^{\sqrt 2}} d x \\ & =\lim _{t \rightarrow \infty} \left[ \frac{x^{-\sqrt {2}+1}}{-\sqrt {2}+1} \right]_{1}^{t}\\ & =\lim _{t \rightarrow \infty} \frac{1}{-\sqrt {2}+1} \left[ \frac{1}{t^{\sqrt {2}-1}}-1 \right] \\ & = \frac{1}{\sqrt {2}-1} \end{split} $$ Thus the given improper integral $$ \int_{1}^{\infty} \frac{1}{x^{\sqrt 2}} d x =\frac{1}{\sqrt {2}-1} $$ So, the integral is convergent.
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