Answer
$$\frac{1}{4} - \frac{{13}}{4}{e^{ - 4}}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {{x^2}{e^{ - 2x}}} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = {x^2} \Rightarrow du = 2xdx; \cr
& dv = {e^{ - 2x}}dx \Rightarrow v = - \frac{1}{2}{e^{ - 2x}} \cr
& {\text{Integration by parts formula }} \cr
& \int {\underbrace {{x^2}}_u} \underbrace {{e^{ - 2x}}dx}_{dv} = \underbrace {{x^2}}_u\underbrace {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)}_v - \int {\underbrace {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)}_v} \underbrace {\left( {2x} \right)dx}_{du} \cr
& \int {{x^2}{e^{ - x}}dx} = - \frac{1}{2}{x^2}{e^{ - 2x}} + \int {x{e^{ - 2x}}} \cr
& \cr
& {\text{Let }}u = x \Rightarrow du = dx; \cr
& dv = {e^{ - 2x}}dx \Rightarrow v = - \frac{1}{2}{e^{ - x}} \cr
& \int {{x^2}{e^{ - x}}dx} = - \frac{1}{2}{x^2}{e^{ - 2x}} + \left( { - \frac{1}{2}x{e^{ - 2x}} - \int {\left( { - \frac{1}{2}{e^{ - 2x}}} \right)dx} } \right) \cr
& \int {{x^2}{e^{ - x}}dx} = - \frac{1}{2}{x^2}{e^{ - 2x}} - \frac{1}{2}x{e^{ - 2x}} + \frac{1}{2}\int {{e^{ - 2x}}dx} \cr
& \int {{x^2}{e^{ - x}}dx} = - \frac{1}{2}{x^2}{e^{ - 2x}} - \frac{1}{2}x{e^{ - 2x}} - \frac{1}{4}{e^{ - 2x}} + C \cr
& \cr
& \int_0^2 {{x^2}{e^{ - 2x}}} dx = \left[ { - \frac{1}{2}{x^2}{e^{ - 2x}} - \frac{1}{2}x{e^{ - 2x}} - \frac{1}{4}{e^{ - 2x}}} \right]_0^2 \cr
& = \left[ { - \frac{1}{2}{{\left( 2 \right)}^2}{e^{ - 4}} - \frac{1}{2}\left( 2 \right){e^{ - 4}} - \frac{1}{4}{e^{ - 4}}} \right] - \left[ { - \frac{1}{4}{e^0}} \right] \cr
& = - 2{e^{ - 4}} - {e^{ - 4}} - \frac{1}{4}{e^{ - 4}} + \frac{1}{4} \cr
& = \frac{1}{4} - \frac{{13}}{4}{e^{ - 4}} \cr} $$
