Answer
$$l = \frac{2}{3}(2\sqrt{2}-1)$$
Work Step by Step
$$y'=x^\frac{1}{2}$$
$$l = \int_{0}^{1}\sqrt{1+{(\sqrt{x})}^2}dx$$
$$l = \int_{0}^{1}\sqrt{1+x}dx$$
$$l = [\frac{2}{3}{(1+x)}^\frac{3}{2}]_{0}^{1}$$
$$l = \frac{2}{3}(2\sqrt{2}-1)$$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 7 - Applications of Integration - 7.4 Exercises - Page 473: 5
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