Single Variable Calculus, Chapter 6, 6.1, Section 6.1, Problem 14

Sketch the region enclosed by the curves $y = \cos x$ , $ y = 2 - \cos x$, $0 \leq x \leq 2 \pi$. Then find the area of the region.


By using vertical strips
$\displaystyle A = \int^{x_2}_{x_1} \left(y_{\text{upper}} - y_{\text{lower}} \right) dx$
In order to get the values of the upper and lower limits, we equate the two functions to get its point of intersection. Thus

$
\begin{equation}
\begin{aligned}
\cos x &= 2 - \cos x\\
\\
2 - 2 \cos x &= 0 \\
\
2( 1- \cos x ) &= 0 \\
\\
\cos x &= 1\\
\\
x & = \cos^{-1}[1]\\
\\
x & = 0 + 2 \pi n \quad \text{;where }n\text{ is any integer}
\end{aligned}
\end{equation}
$

For the interval $ 0 \leq x \leq 2 \pi$,
We have, $x = 0$ and $ x = 2\pi$
Therefore,

$
\begin{equation}
\begin{aligned}
A &= \int^{2 \pi}_0 \left[(2-\cos x) - \cos x \right] dx\\
\\
A &= \int^{2 \pi}_0 \left[ 2-2\cos x \right] dx\\
\\
A &= [ 2 x - 2 \sin x ]^{2\pi}_0\\
\\
A & = 4 \pi \text{ square units}
\end{aligned}
\end{equation}
$