Answer
$$\frac{{e - 1}}{{2{e^2}}}$$
Work Step by Step
$$\eqalign{
& \int_1^{\sqrt 2 } {x{e^{ - {x^2}}}} dx \cr
& u = - {x^2},\,\,\,du = - 2xdx,\,\,\,xdx = - \frac{1}{2}du \cr
& {\text{for }}x = 1,\,\,\,\,\,\,\,\,\,\,u = - {\left( 1 \right)^2}, = - 1 \cr
& {\text{for }}x = \sqrt 2 ,\,\,\,\,\,\,\,\,\,\,u = - {\left( {\sqrt 2 } \right)^2}, = - 2 \cr
& {\text{Thus}}{\text{,}} \cr
& \int_1^{\sqrt 2 } {x{e^{ - {x^2}}}} dx = \int_{ - 1}^{ - 2} {{e^u}\left( { - \frac{1}{2}} \right)} du\, \cr
& = - \frac{1}{2}\int_{ - 1}^{ - 2} {{e^u}} du\, \cr
& {\text{integrating}} \cr
& = - \frac{1}{2}\left[ {{e^u}} \right]_{ - 1}^{ - 2} \cr
& {\text{evaluating the limits}} \cr
& = - \frac{1}{2}\left( {{e^{ - 2}} - {e^{ - 1}}} \right) \cr
& = - \frac{1}{2}\left( {\frac{1}{{{e^2}}} - \frac{1}{e}} \right) \cr
& = \frac{1}{2}\left( {\frac{1}{e} - \frac{1}{{{e^2}}}} \right) \cr
& = \frac{1}{2}\left( {\frac{{e - 1}}{{{e^2}}}} \right) \cr
& = \frac{{e - 1}}{{2{e^2}}} \cr} $$
