If \(\mathrm{f}(x)\) is a function satisfying \(\mathrm{f}^{\prime}(x)=\mathrm{f}(x)\) with \(\mathrm{f}(0)=1\) and \(\mathrm{g}(x)\) is a function that satisfies \(\mathrm{f}(x)+\mathrm{g}(x)=x^2\). Then the value of the integral \(\int_0^1 \mathrm{f}(x) \mathrm{g}(x) \mathrm{d} x\) is

As \(\mathrm{f}^{\prime}(x)=\mathrm{f}(x)\)
\(\frac{f^{\prime}(x)}{\mathrm{f}(x)}=1\)
Integrating on both sides, we get
\(\log \mathrm{f}(x)=x+\mathrm{c}\)
As \(f(0)=1\)
\(\begin{array}{ll}
\therefore & (\mathrm{i}) \Rightarrow \mathrm{c}=0 \\
\therefore & \log \mathrm{f}(x)=x \\
\therefore & \mathrm{f}(x)=\mathrm{e}^x
\end{array}\)
\(\begin{aligned}
& \text { As } \mathrm{f}(x)+\mathrm{g}(x)=x^2 \\
& \mathrm{~g}(x)=x^2-\mathrm{e}^x
\end{aligned}\)
\(\therefore \quad \mathrm{f}(x) \mathrm{g}(x)=\mathrm{e}^x\left(x^2-\mathrm{e}^x\right)\)
\(\begin{aligned}
& =\int_0^1\left(\mathrm{e}^x x^2-\mathrm{e}^{2 x}\right) \mathrm{d} x \\
& =\left[\left(x^2-2 x+2\right) \mathrm{e}^x\right]_0^1-\frac{1}{2} \mathrm{e}^2+\frac{1}{2} \\
& =\mathrm{e}-\frac{1}{2} \mathrm{e}^2-\frac{3}{2}
\end{aligned}\)