Answer
$$A=4 \pi \approx 12.566$$
Also, see Fig.(1)
Work Step by Step
Given $$f(x)=\cos x, g(x)=2-\cos x, 0 \leq x \leq 2 \pi$$
So, the intersection points can be obtained
by equating the two function $f(x)= g(x)$, therefore we get
$$\cos x=2-\cos x, \Rightarrow 2\cos x=2 \Rightarrow \cos x=1 $$$$\Rightarrow x=0 \Rightarrow y=1 \ \ and \ x=2\pi \Rightarrow y=1 $$
First we sketch the region bounded by the graphs $f(x)$ and $ g(x)$ (see Fig.1)
Secondly, we use the integration to find the area of the region
\begin{aligned} A &= \int_{0}^{2 \pi}[g(x)-f(x)] d x\\
&=\int_{0}^{2 \pi}[(2-\cos x)-\cos x] d x \\ &=2 \int_{0}^{2 \pi}(1-\cos x) d x \\ &=2[x-\sin x]_{0}^{2 \pi}=4 \pi \approx 12.566 \end{aligned}
