University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 593: 16


$\dfrac{13}{12} $

Work Step by Step

Since, $L=\int_{1}^{2}\sqrt{1+(\dfrac{dy}{dx})^2}dx$ Thus, $L=\int_{1}^{2} \sqrt{1+\dfrac{1}{16}(y^{4}-\dfrac{1}{2}+\dfrac{1}{y^4})} dy$ or, $L=(\dfrac{1}{2}y^3-y^{-1})_1^{2}$ Thus, $L =\dfrac{13}{12} $
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