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: 12


$$1.91009 $$

Work Step by Step

We have \begin{align*} s&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x\\ &=\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos ^{2}(x)} d x \end{align*} Use the midpoint rule with $n=8$ and $f(x)=\sqrt{1+\cos ^{2}(x)}$ to approximate the arc length, $ \Delta x=\frac{b-a}{n}=\frac{\left(\frac{\pi}{2}-0\right)}{8}=\frac{\pi}{16} $ Then \begin{aligned} s&=\int_{0}^{\frac{\pi}{2}} \sqrt{1+\cos ^{2}(x)} d x\\ &=\Delta x \sum_{i=a}^{8} f\left(x_{i}^{*}\right) \\ & \approx \frac{\pi}{16}[1.411+1.384+1.333+1.264+1.184+1.106+1.041+1.005] \\ & \approx \frac{\pi}{16}[9.728] \\ & \approx 1.91009 \end{aligned}
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