Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 8 - Section 8.1 - Arc Length - 8.1 Exercises - Page 549: 27

Answer

$7.094570$

Work Step by Step

$y = \ln{1+x^{3}}$ then $y = \frac{1}{1+x^{3}} 3x^{2}$ $L = \int^{5}_{0} f(x) dx$ where $f(x) = \sqrt{1+9x^{4}/(1+x^{3})^{2}}$ Since $n = 10, \Delta x = \frac{5-0}{10} = \frac{1}{2}$ $L \approx S_{10} = \frac{1/2}{3}[f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + 2f(2) + 4f(2.5) + 2f(3) + 4f(3.5) + 2f(4) + 4f(4.5) + f(5)] \approx 7.094570$
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