Answer
$$A = \frac{{1 - {e^{ - 1}}}}{2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x{e^{ - {x^2}}},{\text{ }}y = \tan x,{\text{ }}0 \leqslant x \leqslant 1 \cr
& {\text{From the graph shown below, we can define the enclosed area}} \cr
& {\text{as:}} \cr
& A = \int_0^1 {x{e^{ - {x^2}}}dx} \cr
& A = - \frac{1}{2}\int_0^1 {{e^{ - {x^2}}}\left( { - 2x} \right)dx} \cr
& {\text{Integrate and evaluate}} \cr
& A = - \frac{1}{2}\left[ {{e^{ - {x^2}}}} \right]_0^1 \cr
& A = - \frac{1}{2}\left[ {{e^{ - {{\left( 1 \right)}^2}}} - {e^{ - {{\left( 0 \right)}^2}}}} \right] \cr
& A = - \frac{1}{2}\left( {{e^{ - 1}} - 1} \right) \cr
& A = \frac{{1 - {e^{ - 1}}}}{2} \cr} $$
