University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 51


The area of the shaded region is $\pi/2$.

Work Step by Step

The shaded region runs from the point $x=0$ to $x=\pi$. It is bounded above by the line $y=1$ and below by the curve $y=\cos^2x$, So the area of this region would be $$A=\int^\pi_0(1-\cos^2x)dx$$ $$A=\int^\pi_0\sin^2xdx$$ $$A=\int^\pi_0\frac{1-\cos2x}{2}dx$$ $$A=\int^\pi_0\frac{1}{2}dx-\frac{1}{2}\int^\pi_0\cos2xdx$$ $$A=\frac{1}{2}x\Big]^\pi_0-\frac{1}{2}\times\frac{1}{2}\sin2x\Big]^\pi_0$$ $$A=\frac{\pi}{2}-\frac{1}{4}(\sin2\pi-\sin0)$$ $$A=\frac{\pi}{2}-\frac{1}{4}(0-0)=\frac{\pi}{2}$$
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