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