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: 9

Answer

$\frac{2}{243} (82 \sqrt{82} - 1)$

Work Step by Step

$y = 1+6x^{3/2}$ then $dy/dx = 9x^{1/2}$ and $1+(dy/dx)^{2} = 1+81x$ $L = \int^{1}_{0} \sqrt{1+81x} dx = \int ^{82}_{1} u^{1/2}(\frac{1}{81}du) = \frac{1}{81} \frac{2}{3} [u^{3/2}]^{82}_{1} = \frac{2}{243} (82 \sqrt{82} - 1)$
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