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: 7

Answer

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

Work Step by Step

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