Answer
$4\pi$
Work Step by Step
The area is:
$$\begin{align*}
\int_{0}^{2\pi}(2-\cos x-\cos x)dx&=\int_{0}^{2\pi}(2-2\cos x )dx\\
&=[2x-2\sin x ]_{0}^{2\pi}\\
&=4\pi
\end{align*}$$
Calculus 8th Edition
by Stewart, James
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 362: 16
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