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