Calculus 8th Edition

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

Chapter 4 - Integrals - 4.4 Indefinite Integrals and the Net Change Theorem - 4.4 Exercises - Page 339: 70

Answer

$$40$$

Work Step by Step

Given $$\int_{-10}^{10} \frac{2 e^{x}}{\sinh x+\cosh x} d x$$ Since \begin{aligned} \sinh(x)&= \frac{e^x-e^{-x}}{2}\\ \cosh(x)&= \frac{e^x+e^{-x}}{2}\\ \end{aligned} Then \begin{aligned} \sinh(x) + \cosh(x) &= \frac{e^x-e^{-x}}{2}+ \frac{e^x+e^{-x}}{2}\\ & = \frac{2e^x}{2}\\ &=e^x \end{aligned} Hence \begin{aligned} \int_{-10}^{10} \frac{2 e^{x}}{\sinh x+\cosh x} d x&= \int_{-10}^{10} \frac{2 e^{x}}{e^ x} d x\\ &=2\int_{-10}^{10}dx\\ &=2(x)\bigg|_{-10}^{10}\\ &= 2(10-(-10))\\ &= 40 \end{aligned}
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