Calculus: Early Transcendentals 9th Edition

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

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 426: 65

Answer

$e-\sqrt{e}$

Work Step by Step

$\int_{1}^{2}\frac{e^{1/x}}{x^2}~dx$ Let $u = \frac{1}{x}$ $\frac{du}{dx} = -\frac{1}{x^2}$ $dx = -x^2~du$ When $x = 1$, then $u = 1$ When $x = 2$, then $u = \frac{1}{2}$ $\int_{1}^{1/2} \frac{e^u}{x^2}~(-x^2~du)$ $= \int_{1}^{1/2}-e^u~du$ $= \int_{1/2}^{1}e^u~du$ $= e^u~\vert_{1/2}^{1}$ $= e^1-e^{1/2}$ $= e-\sqrt{e}$
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