Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 9 - Further Applications of the Integral and Taylor Polynomials - 9.1 Arc Length and Surface Area - Exercises - Page 468: 8


$$ \frac{17\sqrt{2}}{3}$$

Work Step by Step

Given $$ y=\frac{1}{3}x^{3/2}-x^{1/2}$$ Then arc length given by \begin{aligned} s=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} &=\int_{2}^{8} \sqrt{1+\left(\frac{1}{2} x^{1 / 2}-\frac{1}{2} x^{-1 / 2}\right)^{2}} d x \\ &=\int_{2}^{8} \sqrt{1+\frac{1}{4} x-\frac{1}{2}+\frac{1}{4 x}} d x \\ &=\int_{2}^{8} \sqrt{\frac{1}{2}+\frac{1}{4} x+\frac{1}{4 x}} d x \\ &=\frac{1}{2} \int_{2}^{8} \sqrt{2+x+\frac{1}{x}} d x \\ &=\frac{1}{2} \int_{2}^{8} \frac{x+1}{\sqrt{x}} \\ &=\frac{1}{2} \int_{2}^{8}\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right) d x \\ &=\frac{1}{2}\left[\frac{2}{3} x^{3 / 2}+2 \sqrt{x}\right]_{2}^{8}\\ &= \frac{17\sqrt{2}}{3}\end{aligned}
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