Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 5 - Applications of Integration - 5.1 Areas Between Curves - 5.1 Exercises - Page 363: 31

Answer

$ 4 \over 3$

Work Step by Step

$cos^2x sin x = sin x$ $cos^2x sin x - sin x = 0$ $sin x (cos^2 x -1)=0$ $sin x(-sin^2) = 0$ $sin x = 0$ $x = 0 or \pi$ $A = \int^\pi _0 (sin x - cos^2 x sin x)dx = [-cos x +{{1} \over {3}} cos^2 x]^\pi _0$ $(1-{1 \over 3} )-(-1+{1\over3}) = {4\over3}$
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