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 - Chapter Review Exercises - Page 496: 4

Answer

$$\frac{8}{27}\left[10 \sqrt{10}-\frac{13 \sqrt{13}}{8}\right]$$

Work Step by Step

Given $$y= x^{2/3} ,\ \ \ \ \ \ \ [1,8]$$ Since \begin{align*} y'&= \frac{2}{3}x^{-1/3}\\ 1+y'^2&= 1+ \frac{4}{9}x^{-2/3} \end{align*} Then the arc length is given by \begin{aligned} \text{Arc Length }&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\ &=\int_{1}^{8} \sqrt{1+\frac{4 x^{-2 / 3}}{9}} d x \\ &=\int_{1}^{8} \sqrt{\frac{4 x^{-2 / 3}}{9}\left(1+\frac{9 x^{2 / 3}}{4}\right)} d x \\ &=\int_{1}^{8} \sqrt{\frac{4 x^{-2 / 3}}{9}} \cdot \sqrt{\left(1+\frac{9 x^{2 / 3}}{4}\right)} d x \\ &=\int_{1}^{8} \frac{2 x^{-1 / 3}}{3} \sqrt{1+\frac{9 x^{2 / 3}}{4}} d x \\ &=\frac{8}{27}\left(\frac{9 x^{2 / 3}}{4}+1\right)^{3 / 2}\bigg|_{1}^{8}\\ &= \frac{8}{27}\left[10 \sqrt{10}-\frac{13 \sqrt{13}}{8}\right] \end{aligned}
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