Let f(x) be a function satisfying $f^{\text{'}}\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and $g\left(x\right)$ be a function that satisfies $f\left(x\right)+g\left(x\right)=x^2.$ Then $\underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx=$

$f^{\text{'}}\left(x\right)=f\left(x\right)$, $f\left(0\right)=1$

$$\begin{gathered} f\left(x\right)+g\left(x\right)=x^2 \\ \Rightarrow g\left(x\right)=x^2-f\left(x\right) \\ f\text{'}\left(x\right)=f\left(x\right), f\left(0\right)=1 \\ \Rightarrow f\left(x\right)=e^x \\ \left[{\left(e^x\right)}^{\text{'}}=e^{x}and e^0=1\right] \\ g\left(x\right)=x^2-f\left(x\right)=x^2-e^x \end{gathered}$$

$$\begin{gathered} \underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx \\ =\underset{0}{\overset{1}{\int }}{e}^x\left(x^2-e^x\right)dx \\ =\underset{0}{\overset{1}{\int }}{x}^2.e^{x}dx-\underset{0}{\overset{1}{\int }}{e}^{2x}dx \left[by parts\right] \\ =e^x{\left(x^2-2x+2\right)}_0^1-{\left(\frac{e^{2x}}{2}\right)}_0^1 \\ =e\left(1-2+2\right)-e^0\left(0-0+2\right)-\frac{1}{2}\left(e^2-e^0\right) \\ =e-2-\frac{1}{2}\left(e^2-1\right) \\ =e-\frac{e^2}{2}-\frac{3}{2} \left[x^n{e}^{x}dx=e^x\left(x^n-n.x^{n-1}+n\left(x-1\right)x^{n-2}\right)+......\right] \end{gathered}$$