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 469: 25


$$\frac{a}{2} \sqrt{1+4 a^{2}}+\frac{1}{4} \ln |\sqrt{1+4 a^{2}}+2 a|$$ $$1.478943$$

Work Step by Step

Since \begin{align*} s&=\int_{a}^{b} \sqrt{1+(f'(x))^{2}} d x\\ &= \int_{0}^{a} \sqrt{1+4 x^{2}} d x \end{align*} Let $2x=\tan u ,\ \ 2dx=\sec^2 udu$; then by $$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u$$ we get \begin{align*} s&=\int_{a}^{b} \sqrt{1+(f'(x))^{2}} d x\\ &= \int_{0}^{a} \sqrt{1+4 x^{2}} d x\\ &= \frac{1}{2}\int_{0}^{\tan^{-1}(2a)}\sec^3 udu\\ &=\left.\left(\frac{1}{4} \sec u \tan u+\frac{1}{4} \ln |\sec u+\tan u|\right)\right|_{ 0} ^{\tan^{-1}(2a)}\\ &=\frac{a}{2} \sqrt{1+4 a^{2}}+\frac{1}{4} \ln |\sqrt{1+4 a^{2}}+2 a| \end{align*} Hence, at $a=1$ $$s=\frac{1}{2} \sqrt{5}+\frac{1}{4} \ln (\sqrt{5}+2) \approx 1.478943$$
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