Answer
$$A = 4$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2\sin x + \cos 2x,{\text{ }}y = 0,{\text{ }}0 \leqslant x \leqslant \pi \cr
& {\text{From the graph we can define the area as:}} \cr
& A = \int_0^\pi {\left( {2\sin x + \cos 2x} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - 2\cos x + \frac{1}{2}\sin 2x} \right]_0^\pi \cr
& A = \left[ { - 2\cos \left( \pi \right) + \frac{1}{2}\sin 2\left( \pi \right)} \right] - \left[ { - 2\cos \left( 0 \right) + \frac{1}{2}\sin 2\left( 0 \right)} \right] \cr
& A = \left[ {2 - 0} \right] - \left[ { - 2 - 0} \right] \cr
& A = 2 + 2 \cr
& A = 4 \cr} $$
