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: 21

Answer

$\sqrt{2} + \ln{(1+\sqrt{2})}$

Work Step by Step

$y = \frac{2}{1} x^{2}$ then $y' = x$ and $1+(dy/dx)^{2} = 1 + x^{2}$ So $L = \int^{1}_{-1} \sqrt{1+x^{2}} dx = 2 \int^{1}_{0} \sqrt{1+x^{2}}dx = 2[\frac{x}{2} \sqrt{1+x^{2}} + \frac{1}{2} \ln(x+\sqrt{1+x^{2}})]^{1}_{0}$ $= \sqrt{2} + \ln{1+\sqrt{2}}$
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