Answer
$$ - 39{e^{ - 2}} - {e^2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\left( {{x^3} + 1} \right){e^{ - x}}} dx \cr
& {\text{Integrate by tabulation}} \cr
& \frac{d}{{dx}}:{\text{ }}\left( + \right)\left( {{x^3} + 1} \right),{\text{ }}\left( - \right)3{x^2},{\text{ }}\left( + \right)6x,{\text{ }}\left( - \right)6,{\text{ 0}} \cr
& \int : {\text{ }}{e^{ - x}},{\text{ }} - {e^{ - x}},{\text{ }}{e^{ - x}},{\text{ }} - {e^{ - x}},{\text{ }}{e^{ - x}} \cr
& = \left( {{x^3} + 1} \right)\left( { - {e^{ - x}}} \right) - 3{x^2}\left( {{e^{ - x}}} \right) + 6x\left( { - {e^{ - x}}} \right) - 6\left( {{e^{ - x}}} \right) + C \cr
& = - \left( {{x^3} + 1} \right){e^{ - x}} - 3{x^2}{e^{ - x}} - 6x{e^{ - x}} - 6{e^{ - x}} + C \cr
& \int_{ - 2}^2 {\left( {{x^3} + 1} \right){e^{ - x}}} dx = \left[ { - \left( {{x^3} + 1} \right){e^{ - x}} - 3{x^2}{e^{ - x}} - 6x{e^{ - x}} - 6{e^{ - x}}} \right]_{ - 2}^2 \cr
& = \left[ {\left( { - {x^3} - 1 - 3{x^2} - 6x - 6} \right){e^{ - x}}} \right]_{ - 2}^2 \cr
& = \left[ {\left( { - {x^3} - 3{x^2} - 6x - 7} \right){e^{ - x}}} \right]_{ - 2}^2 \cr
& {\text{Evaluating}} \cr
& {\text{ = }}\left[ {\left( { - {{\left( 2 \right)}^3} - 3{{\left( 2 \right)}^2} - 6\left( 2 \right) - 7} \right){e^{ - 2}}} \right] \cr
& - \left[ {\left( { - {{\left( { - 2} \right)}^3} - 3{{\left( { - 2} \right)}^2} - 6\left( { - 2} \right) - 7} \right){e^2}} \right] \cr
& {\text{Simplifying}} \cr
& {\text{ = }} - 39{e^{ - 2}} - {e^2} \cr} $$
