Answer
$$A = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& y = {\sin ^2}\pi x,{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 1 \cr
& {\text{The area of the region is given by }} \cr
& A = \int_0^1 {{{\sin }^2}\pi x} dx \cr
& A = \int_0^1 {\frac{{1 - \cos 2\pi x}}{2}} dx \cr
& A = \int_0^1 {\left( {\frac{1}{2} - \frac{{\cos 2\pi x}}{{2\left( {2\pi } \right)}}\left( {2\pi } \right)} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{1}{2}x - \frac{1}{{4\pi }}\sin \left( {2\pi x} \right)} \right]_0^1 \cr
& A = \left[ {\frac{1}{2}\left( 1 \right) - \frac{1}{{4\pi }}\sin \left( {2\pi \left( 1 \right)} \right)} \right] - \left[ {\frac{1}{2}\left( 0 \right) - \frac{1}{{4\pi }}\sin \left( 0 \right)} \right] \cr
& A = \left[ {\frac{1}{2}\left( 1 \right) - 0} \right] - \left[ 0 \right] \cr
& A = \frac{1}{2} \cr} $$
