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: 8

Answer

$L = \int ^{1}_{-1} \sqrt{1+4y^2e^{2y^{2}}}dy$

Work Step by Step

$y^{2} = \ln x$ then $x = e^{y^{2}}$ and $dx/dy = 2ye^{y^{2}}$ and $1+(dx/dy)^2 = 1+4y^2e^{2y^{2}}$ $L = \int ^{1}_{-1} \sqrt{1+4y^2e^{2y^{2}}}dy$
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