Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 7 - Applications of Integration - 7.2 Exercises: 35

Answer

$$V=10.0359$$

Work Step by Step

$V=\pi\int_1^2(e^{x-1})^2 dx$ $V=\pi\int_1^2(e^{2x-2})dx$ $v=\pi[\frac{e^{2x-2}}{2}]_1^2$ $V=\frac{\pi}{2} [e^{2x-2}]_1^2$ $V=\frac{\pi}{2}(e ^{2 (2)-2}-e^{2(1)-2})$ $V=\frac{\pi}{2}(e^2-1)$ $V=10.0359$
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