Answer
$$A = \frac{1}{{90}} + \frac{1}{{18}}{e^6}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{{10}}x{e^{3x}},{\text{ }}y = 0,{\text{ }}x = 0,{\text{ }}x = 2 \cr
& {\text{From the graph we can define the area as}} \cr
& A = \int_0^2 {\frac{1}{{10}}x{e^{3x}}} dx \cr
& {\text{Integrating by parts}} \cr
& u = \frac{1}{{10}}x,{\text{ }}du = \frac{1}{{10}}dx \cr
& dv = {e^{3x}}dx.{\text{ }}v = \frac{1}{3}{e^{3x}} \cr
& \int {udv = uv - \int {vdu} } \cr
& \int {\frac{1}{{10}}x{e^{3x}}dx} = \left( {\frac{1}{3}{e^{3x}}} \right)\left( {\frac{1}{{10}}x} \right) - \int {\left( {\frac{1}{3}{e^{3x}}} \right)\left( {\frac{1}{{10}}} \right)dx} \cr
& \int {\frac{1}{{10}}x{e^{3x}}dx} = \frac{1}{{30}}x{e^{3x}} - \frac{1}{{30}}\int {{e^{3x}}dx} \cr
& \int {\frac{1}{{10}}x{e^{3x}}dx} = \frac{1}{{30}}x{e^{3x}} - \frac{1}{{90}}{e^{3x}} + C \cr
& {\text{Therefore}} \cr
& A = \int_0^2 {\frac{1}{{10}}x{e^{3x}}} dx \cr
& A = \left[ {\frac{1}{{30}}x{e^{3x}} - \frac{1}{{90}}{e^{3x}}} \right]_0^2 \cr
& A = \left[ {\frac{1}{{30}}\left( 2 \right){e^{3\left( 2 \right)}} - \frac{1}{{90}}{e^{3\left( 2 \right)}}} \right] - \left[ {\frac{1}{{30}}\left( 0 \right){e^{3\left( 0 \right)}} - \frac{1}{{90}}{e^{3\left( 0 \right)}}} \right] \cr
& A = \frac{1}{{15}}{e^6} - \frac{1}{{90}}{e^6} + \frac{1}{{90}} \cr
& A = \frac{1}{{90}} + \frac{1}{{18}}{e^6} \cr} $$
