Calculus 8th Edition

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

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 4

Answer

$L = \int^{2}_{0}\sqrt{1+e^{-2x}(1-x)^2} dx$

Work Step by Step

$y = x e^{-x}$ then $dy/dx = x(-e^{-x}) + e^{-x}(1) = e^{-x}(1-x)$ and $1+(dy/dx)^2 = 1+[e^{-x}(1-x)]^2$ Therefore $L = \int^{2}_{0}\sqrt{1+e^{-2x}(1-x)^2} dx$
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