Answer
$$2$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {{x^2}{e^{ - x}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& \int_0^\infty {{x^2}{e^{ - x}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{x^2}{e^{ - x}}dx} \cr
& {\text{Integrating by tabulation, you get}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ { - {x^2}{e^{ - x}} - 2x{e^{ - x}} - 2{e^{ - x}}} \right]_0^b \cr
& {\text{Use the Fundamental Theorem of Calculus}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\left( { - {b^2}{e^{ - b}} - 2b{e^{ - b}} - 2{e^{ - b}}} \right) - \left( { - 0{e^0} - 0{e^0} - 2{e^0}} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ { - {b^2}{e^{ - b}} - 2b{e^{ - b}} - 2{e^{ - b}} + 2} \right] \cr
& {\text{Evaluate the limit}} \cr
& = \underbrace { - {{\left( \infty \right)}^2}{e^{ - \infty }}}_0 - \underbrace {2\left( \infty \right){e^{ - \infty }}}_0 - \underbrace {2{e^{ - \infty }}}_0 + 2 \cr
& {\text{Simplify}} \cr
& = 2 \cr} $$
