Question

${\int }_0^1{e}^x{x}^2-e^{2x}dx$

Answer 1

$I={\int }_0^1(e^x{x}^2-e^{2x})dx$

$\therefore$By integrating by parts

$I={\int }_0^1{e}^x{x}^{2}dx-{\int }_0^1{e}^{x}dx$

$I={\left[e^x{(x}^2\right)]}_0^1-{\int }_0^1\left(2x\right)e^{x}dx-{\int }_0^1{e}^{x}dx$

$I=e^1\left(1\right)-0-{\left[\right(2x\left)e^x\right]}_0^1+{\int }_0^{1}2e^{x}dx-{\int }_0^1{e}^{x}dx$

$I=e-2e+2e-2e^0-\frac{1}{2}{[e^2-e^0]}^0$

$I=e-2-\frac{e^2}{2}+\frac{1}{2}$

$I=e-\frac{e^2}{2}-\frac{3}{2}$