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: 28

Answer

$2.280559$

Work Step by Step

$y = e^{-x^{2}}$ then $dy/dx = e^{-x^{2}} (-2x)$ $L = \int^{2}_{0} f(x) dx$ where $f(x) = \sqrt{1+4x^{2} e^{-2x^{2}}}$ Since $n = 10, \Delta x = \frac{2-0}{10} = \frac{1}{5}$ $L \approx S_{10} = \frac{1/5}{3} [f(0) + 4f(0.2) + 2f(0.4) + 4f(0.6) + 2f(0.8) + 4f(1) + 2f(1.2) + 4f(1.4) + 2f(1.6) + 4f(1.8) + f(2)] \approx 2.280559$
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