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 - Page 589: 5


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

Work Step by Step

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