Answer
$$1$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {x{e^{ - x}}dx} \cr
& {\text{definition of improper integral}} \cr
& \int_0^\infty {x{e^{ - x}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {x{e^{ - x}}dx} \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( { - x{e^{ - x}} - {e^{ - x}}} \right)} \right|_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - b{e^{ - b}} - {e^{ - b}} - \left( { - 0{e^0} - {e^0}} \right)} \right) \cr
& {\text{simplify}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( { - b{e^{ - b}} - {e^{ - b}} + 1} \right) \cr
& {\text{evaluate the limit}} \cr
& = - \infty {e^{ - \infty }} - {e^{ - \infty }} + 1 \cr
& = 0 - 0 + 1 \cr
& = 1 \cr} $$
