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: 31



Work Step by Step

Since $0 \leq f^{\prime}(x) \leq 1,$ then $$ \sqrt{1+\left(f^{\prime}(x)\right)^{2}} \leq \sqrt{1+1}=\sqrt{2} $$ By comparison, we have $$ \int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x \leq \int_{a}^{b} \sqrt{2} d x=\sqrt{2}(b-a) $$ For $f(x)=x,$ we have $$ f^{\prime}(x)=1 $$ Therefore, we get $$ \int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x=\int_{a}^{b} \sqrt{2} d x=\sqrt{2}(b-a) $$
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