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 $1+(dx/dy)^2 = 1+4y^2e^{2y^{2}}$
$L = \int ^{1}_{-1} \sqrt{1+4y^2e^{2y^{2}}}dy$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 8
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