Answer
$$A = 2 - 8{e^{ - 3}}$$
Work Step by Step
$$\eqalign{
& y = 2x{e^{ - x}},{\text{ }}y = 0,{\text{ }}x = 3 \cr
& {\text{From the graph we can define the area as}} \cr
& A = \int_0^3 {2x{e^{ - x}}} dx \cr
& {\text{Integrating by parts}} \cr
& u = 2x,{\text{ }}du = 2dx \cr
& dv = {e^{ - x}}dx.{\text{ }}v = - {e^{ - x}} \cr
& \int {udv = uv - \int {vdu} } \cr
& \int {2x{e^{ - x}}dx} = \left( {2x} \right)\left( { - {e^{ - x}}} \right) - \int {\left( { - {e^{ - x}}} \right)\left( 2 \right)dx} \cr
& \int {2x{e^{ - x}}dx} = - 2x{e^{ - x}} + 2\int {{e^{ - x}}dx} \cr
& \int {2x{e^{ - x}}dx} = - 2x{e^{ - x}} - 2{e^{ - x}} + C \cr
& {\text{Therefore}} \cr
& A = \int_0^3 {2x{e^{ - x}}} dx \cr
& A = \left[ { - 2x{e^{ - x}} - 2{e^{ - x}}} \right]_0^3 \cr
& A = \left[ { - 2\left( 3 \right){e^{ - 3}} - 2{e^{ - 3}}} \right] - \left[ { - 2\left( 0 \right){e^{ - 3}} - 2{e^0}} \right] \cr
& A = - 6{e^{ - 3}} - 2{e^{ - 3}} + 2 \cr
& A = 2 - 8{e^{ - 3}} \cr} $$
