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: 3



Work Step by Step

Given $$ y= y=\frac{1}{12} x^{3}+x^{-1}$$ Since \begin{aligned} 1+\left(y^{\prime}\right)^{2} &=1+\left(\frac{1}{4} x^{2}-x^{-2}\right)^{2} \\ &=1+\frac{1}{4} x^{4}-\frac{1}{2}+x^{-4} \\ &=\frac{1}{4} x^{4}+\frac{1}{2}+x^{-4} \\ &=\left(\frac{1}{4} x^{2}+x^{-2}\right)^{2} \end{aligned} Then the arc length given by \begin{aligned} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{1}^{2} \sqrt{\left(\frac{1}{4} x^{2}+x^{-2}\right)^{2}} d x \\ &=\int_{1}^{2}\left(\frac{1}{4} x^{2}+x^{-2}\right) d x \\ &=\left[\frac{1}{12} x^{3}-x^{-1}\right]_{1}^{2} \\ &=\frac{1}{6}+\frac{11}{12} \\ &=\frac{13}{12} \end{aligned}
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