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: 1

Answer

$4\sqrt{5}$

Work Step by Step

$y = 2x-5$ $L = \int^{3}_{-1} \sqrt{1+(\frac{dy}{dx})^2}dx = \int^{3}_{-1}\sqrt{1+(2)^2}dx = \sqrt{5}[3-(-1)] = 4\sqrt{5} $
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