Answer
$L = \int^{4}_{1} \sqrt{1+(1-1/x)^{2}} dx$
Work Step by Step
$x = y^{2} - 2y$ then $dx/ dy = 2y - 2$ and $1+(dx/dy)^2 = 1+(2y-2)^{2}$
Therefore
$L = \int^{4}_{1} \sqrt{1+(1-1/x)^{2}} dx$
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: 6
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