Answer
$$A = \frac{3}{2} - \frac{1}{e}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {e^{ - x}},{\text{ }}g\left( x \right) = - x \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left( {{e^{ - x}} + x} \right)} dx \cr
& {\text{ Integrate}} \cr
& A = \left[ { - {e^{ - x}} + \frac{1}{2}{x^2}} \right]_0^1 \cr
& A = \left[ { - {e^{ - 1}} + \frac{1}{2}{{\left( 1 \right)}^2}} \right] - \left[ { - {e^0} + \frac{1}{2}{{\left( 0 \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& A = - {e^{ - 1}} + \frac{1}{2}{\left( 1 \right)^2} + 1 \cr
& A = \frac{3}{2} - \frac{1}{e} \cr} $$
