Answer
$$ - x{e^{ - x}} - 2{x^2}{e^{ - x}} - 4{e^{ - x}}\left( {x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{x^3} - {x^2}} \right){e^{ - x}}} dx \cr
& {\text{use the distributive property}} \cr
& = \int {{x^3}{e^{ - x}}} dx - \int {{x^2}{e^{ - x}}} dx \cr
& = - \int {{x^3}{e^{ - x}}} \left( { - 1} \right)dx + \int {{x^2}{e^{ - x}}\left( { - 1} \right)} dx \cr
& {\text{let }}u = - x,\,\,\,du = \left( { - 1} \right)dx \cr
& = \int {{u^3}{e^u}} du + \int {{u^2}{e^u}} du \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By the formula 52}} \cr
& \left( {52} \right):\,\,\,\,\,\int {{u^n}{e^u}} du = {u^n}{e^u} - n\int {{u^{n - 1}}{e^u}du} \cr
& = {u^3}{e^u} - 3\int {{u^{3 - 1}}{e^u}du} + {u^2}{e^u} - 2\int {{u^{2 - 1}}{e^u}du} \cr
& = {u^3}{e^u} - 3\int {{u^2}{e^u}du} + {u^2}{e^u} - 2\int {u{e^u}du} \cr
& = {u^3}{e^u} - 3\left( {{u^2}{e^u} - 2\int {u{e^u}du} } \right) + {u^2}{e^u} - 2\int {u{e^u}du} \cr
& = {u^3}{e^u} - 3{u^2}{e^u} + 6\int {u{e^u}du} + {u^2}{e^u} - 2\int {u{e^u}du} \cr
& = {u^3}{e^u} - 2{u^2}{e^u} + 4\int {u{e^u}du} \cr
& {\text{By the tables use the formula }}\int {u{e^u}du} = {e^u}\left( {u - 1} \right) + C \cr
& = {u^3}{e^u} - 2{u^2}{e^u} + 4{e^u}\left( {u - 1} \right) + C \cr
& \cr
& {\text{Write in terms of }}x{\text{ replace }} - x{\text{ for }}u \cr
& = {\left( { - x} \right)^3}{e^{ - x}} - 2{\left( { - x} \right)^2}{e^{ - x}} + 4{e^{ - x}}\left( { - x - 1} \right) + C \cr
& = - x{e^{ - x}} - 2{x^2}{e^{ - x}} - 4{e^{ - x}}\left( {x + 1} \right) + C \cr} $$
