Chapter 8 - Further Techniques and Applications of Integration - 8.1 Integration by Parts - 8.1 Exercises - Page 433: 19

$$ - \frac{{3{e^4} + {e^2}}}{4}$$

$$\eqalign{ & \int_1^2 {\left( {1 - {x^2}} \right)} {e^{2x}}dx \cr & {\text{setting }}\,\,\,\,\,\,u = 1 - {x^2}{\text{ then }}du = - 2xdx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{2x}}dx{\text{ then }}v = \frac{1}{2}{e^{2x}} \cr & {\text{Substituting these values into the formula for integration by parts}} \cr & \int u dv = uv - \int {vdu} \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} - \int {\left( {\frac{1}{2}{e^{2x}}} \right)} \left( { - 2xdx} \right) \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \int {x{e^{2x}}} dx \cr & \cr & {\text{integrate by parts again }}\int {x{e^{2x}}} dx \cr & {\text{setting }}\,\,\,\,\,\,u = x{\text{ then }}du = dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {e^{2x}}dx{\text{ then }}v = \frac{1}{2}{e^{2x}} \cr & {\text{then}} \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \left( {\frac{x}{2}{e^{2x}} - \int {\frac{1}{2}{e^{2x}}dx} } \right) \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \frac{x}{2}{e^{2x}} - \int {\frac{1}{2}{e^{2x}}dx} \cr & {\text{integrate}} \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \frac{x}{2}{e^{2x}} - \frac{1}{2}\left( {\frac{{{e^{2x}}}}{2}} \right) + C \cr & \int {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \frac{x}{2}{e^{2x}} - \frac{{{e^{2x}}}}{4} + C \cr & \cr & {\text{Now find the definite integral}} \cr & \int_1^2 {\left( {1 - {x^2}} \right)} {e^{2x}}dx = \left[ {\frac{1}{2}\left( {1 - {x^2}} \right){e^{2x}} + \frac{x}{2}{e^{2x}} - \frac{{{e^{2x}}}}{4}} \right]_1^2 \cr & = \left[ {\frac{1}{2}\left( {1 - {2^2}} \right){e^{2\left( 2 \right)}} + \frac{2}{2}{e^{2\left( 2 \right)}} - \frac{{{e^{2\left( 2 \right)}}}}{4}} \right] - \left[ {\frac{1}{2}\left( {1 - {1^2}} \right){e^{2\left( 1 \right)}} + \frac{1}{2}{e^{2\left( 1 \right)}} - \frac{{{e^{2\left( 1 \right)}}}}{4}} \right] \cr & {\text{simplifying}} \cr & = \left[ { - \frac{3}{2}{e^4} + {e^4} - \frac{1}{4}{e^4}} \right] - \left[ {0 + \frac{1}{2}{e^2} - \frac{{{e^2}}}{4}} \right] \cr & = - \frac{3}{4}{e^4} - \frac{1}{4}{e^2} \cr & = - \frac{{3{e^4} + {e^2}}}{4} \cr} $$