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: 6


$L = \int^{4}_{1} \sqrt{1+(1-1/x)^{2}} dx$

Work Step by Step

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