Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.7 Improper Integrals - Exercises - Page 442: 82

Answer

$\dfrac{\pi}{2}$

Work Step by Step

We calculate the volume as follows: Volume $=\int_0^{\infty} \pi (e^{-x})^2 \ dx\\=\pi \int_0^{\infty} (e^{-x})^2 \ dx \\=\pi \lim\limits_{R \to \infty}\dfrac{e^{-2x}}{-2}|_0^R\\=\dfrac{\pi}{2} \lim\limits_{R \to \infty} (1-e^{-2R})\\=\dfrac{\pi}{2}$
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