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: 15


$\dfrac{285}{8} $

Work Step by Step

Since, $L=\int_{1}^{32}\sqrt{1+(\dfrac{dy}{dx})^2}dx$ Thus, $L=\int_{1}^{32} \sqrt{1+\dfrac{1}{4}(x^{2/5}-2+x^{-2/5}} dx=\int_{1}^{32} (x^{1/5}+x^{-1/5})dx$ or, $L=(\dfrac{1}{2})[(5/6)x^{6/5}+(5/4)x^{4/5}]_1^{32}$ Thus, $L =\dfrac{285}{8} $
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