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: 6

Answer

$$2\sqrt{10}.$$

Work Step by Step

To find the arc length $s$ of the curve $y=9-3x$ between $x=1$ and $x=3$, we fist calculate $y'=-3$. Then we have the integral, $$s=\int_1^{3}\sqrt{1+(y')^2}dx=\int_1^{3}\sqrt{1+9}dx=\sqrt{10}x|_1^3=2\sqrt{10}.$$
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