## Calculus with Applications (10th Edition)

$$10\sqrt e - 16$$
\eqalign{ & \int_0^1 {{x^2}{e^{x/2}}} dx \cr & {\text{setting }}\,\,\,\,\,\,u = {x^2}{\text{ then }}du = 2xdx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{x/2}}dx{\text{ then }}v = 2{e^{x/2}} \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - \int {\left( {2{e^{x/2}}} \right)} \left( {2xdx} \right) \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - 4\int {x{e^{x/2}}} dx \cr & \cr & {\text{integrate by parts again }}\int {x{e^{x/2}}} dx \cr & {\text{setting }}\,\,\,\,\,\,u = x{\text{ then }}du = dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{x/2}}dx{\text{ then }}v = 2{e^{x/2}} \cr & \cr & {\text{then}} \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - 4\left( {2x{e^{x/2}} - \int {2{e^{x/2}}dx} } \right) \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - 8x{e^{x/2}} + 4\int {2{e^{x/2}}dx} \cr & {\text{integrate}} \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - 8x{e^{x/2}} + 8\left( {2{e^{x/2}}} \right) + C \cr & \int {{x^2}} {e^{x/2}}dx = 2{x^2}{e^{x/2}} - 8x{e^{x/2}} + 16{e^{x/2}} + C \cr & \cr & {\text{then}} \cr & \int_0^1 {{x^2}{e^{x/2}}} dx = \left[ {2{x^2}{e^{x/2}} - 8x{e^{x/2}} + 16{e^{x/2}}} \right]_0^1 \cr & {\text{evaluating the limits}} \cr & = \left[ {2{{\left( 1 \right)}^2}{e^{1/2}} - 8\left( 1 \right){e^{1/2}} + 16{e^{1/2}}} \right] - \left[ {\frac{{{{\left( 0 \right)}^2}{e^{1/2}}}}{2} - 8\left( 0 \right){e^{1/2}} + 16{e^{0/2}}} \right] \cr & {\text{simplifying}} \cr & = \left[ {2{e^{1/2}} - 8{e^{1/2}} + 16{e^{1/2}}} \right] - \left[ {\frac{{{{\left( 0 \right)}^2}{e^{1/2}}}}{2} - 8\left( 0 \right){e^{1/2}} + 16{e^{1/2}}} \right] \cr & = 10{e^{1/2}} - 16 \cr & = 10\sqrt e - 16 \cr}