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Area intercepted by the curves $y=\cos x$ $\mathrm{x} \in[0, \pi]$ and $\mathrm{y}=\cos 2 \mathrm{x}, \mathrm{x} \in[0, \pi],$ is
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The correct answer is:
$\frac{3 \sqrt{3}}{4}$
$$
\text { Area }=\int_{0}^{2 \pi / 3}(\cos x-\cos 2 x) d x=\frac{3 \sqrt{3}}{4}
$$
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