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: 98


$V= \pi (1-e^{-1})=1.986$

Work Step by Step

The volume of the solid obtained by rotating the region under the curve $y=e^{-x^{2}}$ From 0 to 1 about the x-axis is equal to $V=\int_{0}^{1} 2\pi xe^{-x^{2}} dx$ Consider $x^{2}=t$ $2xdx=dt$ $dx=\frac{1}{2x}dt$ Then $V=\pi \int_{0}^{1} e^{-t} dt$ $V=\pi (e^{0}-e^{-1})$ Hence, $V= \pi (1-e^{-1})=1.986$
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