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: 35

Answer

$\frac{2}{27}[(1+9x)^{3/2} - 10\sqrt{10}]$

Work Step by Step

$y = 2x^{3/2}$ then $y' = 3x^{1/2}$ and $1+(y')^{2} = 1+9x$ Starting at $P_0 (1, 2)$ $s(x) = \int^{x}_{1} \sqrt{1+9t} dt = [\frac{2}{27} (1+9t)^{3/2}]^{x}_{1} = \frac{2}{27}[(1+9x)^{3/2} - 10\sqrt{10}]$
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