Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Review - Exercises - Page 423: 44

Answer

$Area = 0$

Work Step by Step

$f(x) = cos^2~x~sin~x$ Under the graph on the interval $0 \leq x \leq 2\pi$, we can see that there is symmetry about the x-axis. That is, half the area is above the x-axis and half the area is below the x-axis. Therefore, the area is 0. We can find the exact area $A$ under the graph: $A = \int_{0}^{2\pi}cos^2~x~sin~x~dx$ $A = -\frac{1}{3}cos^3~x\Big\vert_{0}^{2\pi}$ $A = -\frac{1}{3}cos^3~(2\pi) - [-\frac{1}{3}cos^3~(0)]$ $A = -\frac{1}{3}(1)^3 +\frac{1}{3}(1)^3$ $A = 0$
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