Answer
$$A = \frac{\pi }{{{\pi ^2} + 1}}\left( {\frac{1}{e} + 1} \right)$$
Work Step by Step
$$\eqalign{
& y = {e^{ - x}}\sin \pi x,{\text{ }}y = 0,{\text{ }}x = 1 \cr
& {\text{From the graph we can define the area as}} \cr
& A = \int_0^1 {{e^{ - x}}\sin \pi x} dx \cr
& {\text{Integrating by the formula of exercise 71}} \cr
& \int {{e^{ax}}\sin bxdx} = \frac{{{e^{ax}}\left( {a\sin bx - b\cos bx} \right)}}{{{a^2} + {b^2}}} + C \cr
& A = \int_0^1 {{e^{ - x}}\sin \pi x} dx \cr
& a = - 1,{\text{ }}b = \pi \cr
& A = \left[ {\frac{{{e^{ - x}}\left( { - \sin \pi x - \pi \cos \pi x} \right)}}{{{{\left( { - 1} \right)}^2} + {{\left( \pi \right)}^2}}}} \right]_0^1 \cr
& A = - \left[ {\frac{{{e^{ - x}}\left( {\sin \pi x + \pi \cos \pi x} \right)}}{{{\pi ^2} + 1}}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& A = - \left[ {\frac{{{e^{ - 1}}\left( {\sin \pi + \pi \cos \pi } \right)}}{{{\pi ^2} + 1}}} \right] + \left[ {\frac{{{e^0}\left( {\sin 0 + \pi \cos 0} \right)}}{{{\pi ^2} + 1}}} \right] \cr
& A = - \left[ {\frac{{{e^{ - 1}}\left( { - \pi } \right)}}{{{\pi ^2} + 1}}} \right] + \left[ {\frac{\pi }{{{\pi ^2} + 1}}} \right] \cr
& A = \frac{{{e^{ - 1}}\pi }}{{{\pi ^2} + 1}} + \frac{\pi }{{{\pi ^2} + 1}} \cr
& A = \frac{{\pi \left( {{e^{ - 1}} + 1} \right)}}{{{\pi ^2} + 1}} \cr
& A = \frac{\pi }{{{\pi ^2} + 1}}\left( {\frac{1}{e} + 1} \right) \cr} $$
