Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises: 94

Answer

$\frac{4\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$

Work Step by Step

Let $u=1+e^{-x}$. Then $du=-e^{-x}=-\frac{1}{e^x}$, and $-du=\frac{1}{e^x}$. $\int_0^1\frac{\sqrt{1+e^{-x}}}{e^x}dx$ $=\int_{1+e^{-0}}^{1+e^{-1}}\sqrt{u}*(-1)du$ $=-\int_2^{1+\frac{1}{e}}u^{\frac{1}{2}}du$ $=-\left.\left(\frac{u^\frac{3}{2}}{\frac{3}{2}}\right)\right|_2^{1+\frac{1}{e}}$ $=-\left.\left(\frac{2u^\frac{3}{2}}{3}\right)\right|_2^{1+\frac{1}{e}}$ $=-\left(\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}-\frac{2*2^\frac{3}{2}}{3}\right)$ $=\frac{2*2^\frac{3}{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$ $=\frac{2*2\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$ $=\frac{4\sqrt{2}}{3}-\frac{2(1+\frac{1}{e})^\frac{3}{2}}{3}$
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