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
$L = \int^{2}_{0}\sqrt{1+e^{-2x}(1-x)^2} dx$
Calculus 8th Edition
by Stewart, James
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 - Page 589: 4
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