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: 1



Work Step by Step

Given $$y=\frac{x^{5}}{10}+\frac{x^{-3}}{6},\ \ \ [1,2]$$ Since \begin{aligned} y'&= \frac{x^{4}}{2}-\frac{1}{2 x^{4}}\\ \end{aligned} Then the arc length is given by \begin{aligned} s&=\int_{a}^{b} \sqrt{1+y'^{2}} d x \\ &=\int_{1}^{2} \sqrt{1+\frac{x^{8}}{4}-\frac{1}{2}+\frac{1}{4 x^{8}} } d x \\ &=\int_{1}^{2} \sqrt{\frac{x^{8}}{4}+\frac{1}{2}+\frac{1}{4 x^{8}}} d x \\ &=\int_{1}^{2} \sqrt{\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right)^{2}} d x \\ &=\int_{1}^{2} \sqrt{\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right)^{2}} d x \\ &=\int_{1}^{2}\left(\frac{x^{4}}{2}+\frac{1}{2 x^{4}}\right) d x \\ &=\frac{1}{2} \int_{1}^{2}\left(x^{4}+x^{-4}\right) d x \\ &=\frac{1}{2}\left[\frac{x^{5}}{5}-\frac{x^{-3}}{3}\right]_{1}^{2}\\ &= \frac{779}{240} \end{aligned}
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