## Calculus, 10th Edition (Anton)

$$\frac{3}{4}{e^4} + \frac{1}{4}$$
\eqalign{ & \int_0^2 {x{e^{2x}}} dx \cr & {\text{substitute }}u = x,{\text{ }}du = dx \cr & dv = {e^{2x}}dx,{\text{ }}v = \frac{1}{2}{e^{2x}} \cr & {\text{applying integration by parts}}{\text{, we have}} \cr & \int_0^2 {x{e^{2x}}} dx = \left. {\left( {\frac{1}{2}x{e^{2x}}} \right)} \right|_0^2 - \int_0^2 {\left( {\frac{1}{2}{e^{2x}}} \right)dx} \cr & \int_0^2 {x{e^{2x}}} dx = \left. {\left( {\frac{1}{2}x{e^{2x}}} \right)} \right|_0^2 - \frac{1}{2}\int_0^2 {{e^{2x}}dx} \cr & {\text{integrating}} \cr & \int_0^2 {x{e^{2x}}} dx = \left. {\left( {\frac{1}{2}x{e^{2x}}} \right)} \right|_0^2 - \frac{1}{4}\left. {\left( {{e^{2x}}} \right)} \right|_0^2 \cr & \int_0^2 {x{e^{2x}}} dx = \left. {\left( {\frac{1}{2}x{e^{2x}} - \frac{1}{4}{e^{2x}}} \right)} \right|_0^2 \cr & {\text{evaluate limits}} \cr & = \left( {\frac{1}{2}\left( 2 \right){e^{2\left( 2 \right)}} - \frac{1}{4}{e^{2\left( 2 \right)}}} \right) - \left( {\frac{1}{2}\left( 0 \right){e^{2\left( 0 \right)}} - \frac{1}{4}{e^{2\left( 0 \right)}}} \right) \cr & {\text{simplify}} \cr & = \left( {{e^4} - \frac{1}{4}{e^4}} \right) - \left( { - \frac{1}{4}} \right) \cr & = \frac{3}{4}{e^4} + \frac{1}{4} \cr}