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: 16

Answer

$\frac{1}{2} \sinh 2$

Work Step by Step

$y = 3+ \frac{1}{2} \cosh 2x$ then $y' = \sinh 2x$ and $1+(dy/dx)^{2} = 1 + \sinh^{2} 2x = \cosh^{2} 2x$ So $L = \int^{1}_{0} \sqrt{\cosh^{2} 2x} dx = \int^{1}_{0} \cosh 2x dx = [\frac{1}{2} \sinh 2x]^{1}_{0} = \frac{1}{2} \sinh 2 - 0 =\frac{1}{2} \sinh 2$
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