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: 28


$$s=\int_{a}^{b}\sqrt{ \frac{f^2(x)+f'^2(x)}{f^2(x)} }$$

Work Step by Step

Since $$y=\ln [f(x)]\to y'=\frac{f'(x)}{f(x)} $$ Then \begin{align*} s&=\int_{a}^{b}\sqrt{1+y'^2}dx\\ &= \int_{a}^{b}\sqrt{1+\frac{f'^2(x)}{f^2(x)} }\\ &=\int_{a}^{b}\sqrt{ \frac{f^2(x)+f'^2(x)}{f^2(x)} }\\ \end{align*}
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