Answer
$$\frac{{5{e^6} + 1}}{9}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {x{e^{3x}}dx} \cr
& = \frac{1}{9}\int_0^2 {3x{e^{3x}}\left( 3 \right)dx} \cr
& {\text{Integrate by tables, use }}\int {u{e^u}du} = \left( {u - 1} \right){e^u} + C \cr
& \frac{1}{9}\int_0^2 {3x{e^{3x}}\left( 3 \right)dx} = \frac{1}{9}\left[ {\left( {3x - 1} \right){e^{3x}}} \right]_0^2 \cr
& {\text{Evaluate}} \cr
& \frac{1}{9}\left[ {\left( {3x - 1} \right){e^{3x}}} \right]_0^2 = \frac{1}{9}\left[ {\left( {3\left( 2 \right) - 1} \right){e^{3\left( 2 \right)}} - \left( {3\left( 0 \right) - 1} \right){e^{3\left( 0 \right)}}} \right] \cr
& = \frac{1}{9}\left[ {5{e^6} + 1} \right] \cr
& = \frac{{5{e^6} + 1}}{9} \cr} $$
