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: 11

Answer

$\frac{59}{24}$

Work Step by Step

$y = \frac{x^{3}}{3} + \frac{1}{4x}$ then $y' = x^{2} - \frac{1}{4x^{2}}$ $1+(y')^{2} = 1 + (x^{4} - \frac{1}{2} + \frac{1}{16x^{4}}) = x^{4} + \frac{1}{2} + \frac{1}{16x^{4}} = (x^{2} + \frac{1}{4x^2})^{2}$ $L = \int^{2}_{1} \sqrt{1+(y')^{2}} dx = \int ^{2}_{1} (x^{2} + \frac{1}{4x^{2}}) dx = [\frac{1}{3} x^{3} - \frac{1}{4x}]^{2}_{1} = \frac{59}{24}$
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