If $\mathrm{f}(x)$ is a function satisfying $\mathrm{f}^{\prime}(x)=\mathrm{f}(x)$ with $\mathrm{f}(0)=1$…
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
$e-\frac{e^2}{2}-\frac{5}{2}$
$\mathrm{e}+\frac{\mathrm{e}^2}{2}-\frac{3}{2}$
$e-\frac{e^2}{2}-\frac{3}{2}$
$e+\frac{e^2}{2}+\frac{5}{2}$
Solution
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 $f(0)=1$
$\begin{array}{ll}
\therefore & (\mathrm{i}) \Rightarrow \mathrm{c}=0 \\
\therefore & \log \mathrm{f}(x)=x \\
\therefore & \mathrm{f}(x)=\mathrm{e}^x
\end{array}$
$\begin{aligned}
& \text { As } \mathrm{f}(x)+\mathrm{g}(x)=x^2 \\
& \mathrm{~g}(x)=x^2-\mathrm{e}^x
\end{aligned}$
$\therefore \quad \mathrm{f}(x) \mathrm{g}(x)=\mathrm{e}^x\left(x^2-\mathrm{e}^x\right)$
$\begin{aligned}
& =\int_0^1\left(\mathrm{e}^x x^2-\mathrm{e}^{2 x}\right) \mathrm{d} x \\
& =\left[\left(x^2-2 x+2\right) \mathrm{e}^x\right]_0^1-\frac{1}{2} \mathrm{e}^2+\frac{1}{2} \\
& =\mathrm{e}-\frac{1}{2} \mathrm{e}^2-\frac{3}{2}
\end{aligned}$