Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 8 - Further Applications of Integration - 8.1 Arc Length - 8.1 Exercises: 26

Answer

$\approx 5.074212$

Work Step by Step

$y = x^{3/2}$ then $y' = \frac{1}{3} x^{-2/3}$ $L = \int^{6}_{1} f(x) dx$ where $f(x) = \sqrt{1+\frac{1}{9} x^{-4/3}}$ Since $n = 10, \Delta x = \frac{6-1}{10} = \frac{1}{2}$ Now$L = \approx S_{10} = \frac{1/2}{3} [f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + 2f(3) + 4f(3.5) + 2f(4) + 4f(4.5) + 2f(5) + 4f(5.5) + f(6)] \approx 5.074212$
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