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: 17

Answer

$\frac{3}{4} + \frac{1}{2} \ln 2$

Work Step by Step

$y = \frac{1}{4} x^{2} - \frac{1}{2} \ln x$ then $y' = \frac{1}{2} x - \frac{1}{2x}$ and $1+(y')^{2} = 1+(\frac{1}{4} x^{2} -\frac{1}{2} + \frac{1}{4x^{2}}) = \frac{1}{4} x^{2} + \frac{1}{2} + \frac{1}{4x^{2}} = (\frac{1}{2} x + \frac{1}{2x})^{2}$ So $L = \int ^{2}_{1} \sqrt{1+(y')^{2}}dx = \int ^{2}_{1} |\frac{1}{2}x + \frac{1}{2x}| dx = \int ^{2}_{1} (\frac{1}{2}x + \frac{1}{2x})dx = [\frac{1}{4}x^{2} + \frac{1}{2} \ln|x|]^{2}_{1} = \frac{3}{4} + \frac{1}{2} \ln 2$
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