Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.2 Exponential Functions and Their Derivatives - 6.2 Exercises - Page 420: 93



Work Step by Step

Let $u=\frac{1}{x}$. Then $u=x^{-1}$, and $du=-1x^{-2}=-\frac{1}{x^2}$, so $-du=\frac{1}{x^2}$. $\int_1^2\frac{e^{1/x}}{x^2}dx$ $=\int_{\frac{1}{1}}^{\frac{1}{2}}e^u*(-1) du$ $=-\int_1^{\frac{1}{2}}e^udu$ $=-e^u|_1^\frac{1}{2}$ $=-(e^\frac{1}{2}-e^1)$ $=-(\sqrt{e}-e)$ $=e-\sqrt{e}$
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