Answer
$$ - \frac{6}{{25{e^5}}} + \frac{1}{{25}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x{e^{ - 5x}}} dx \cr
& {\text{substitute }}u = x,{\text{ }}du = dx \cr
& dv = {e^{ - 5x}}dx,{\text{ }}v = - \frac{1}{5}{e^{ - 5x}} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_0^1 {x{e^{ - 5x}}} dx = \left. {\left( { - \frac{1}{5}x{e^{ - 5x}}} \right)} \right|_0^1 - \int_0^1 {\left( { - \frac{1}{5}{e^{ - 5x}}} \right)dx} \cr
& \int_0^1 {x{e^{ - 5x}}} dx = \left. {\left( { - \frac{1}{5}x{e^{ - 5x}}} \right)} \right|_0^1 + \frac{1}{5}\int_0^1 {{e^{ - 5x}}dx} \cr
& {\text{integrating}} \cr
& \int_0^1 {x{e^{ - 5x}}} dx = \left. {\left( { - \frac{1}{5}x{e^{ - 5x}}} \right)} \right|_0^1 - \frac{1}{{25}}\left. {\left( {{e^{ - 5x}}} \right)} \right|_0^1 \cr
& \int_0^1 {x{e^{ - 5x}}} dx = \left. {\left( { - \frac{1}{5}x{e^{ - 5x}} - \frac{1}{{25}}{e^{ - 5x}}} \right)} \right|_0^1 \cr
& {\text{evaluate limits}} \cr
& = \left( { - \frac{1}{5}\left( 1 \right){e^{ - 5\left( 1 \right)}} - \frac{1}{{25}}{e^{ - 5\left( 1 \right)}}} \right) - \left( { - \frac{1}{5}\left( 0 \right){e^{ - 5\left( 0 \right)}} - \frac{1}{{25}}{e^{ - 5\left( 0 \right)}}} \right) \cr
& {\text{simplify}} \cr
& = \left( { - \frac{1}{5}{e^{ - 5}} - \frac{1}{{25}}{e^{ - 5}}} \right) - \left( { - \frac{1}{{25}}} \right) \cr
& = - \frac{6}{{25{e^5}}} + \frac{1}{{25}} \cr} $$
