Let $f(x)$ be a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ be a function that…
Let $f(x)$ be a function satisfying $f^{\prime}(x)=f(x)$ with $f(0)=1$ and $g(x)$ be a function that satisfies $f(x)+g(x)=x^2$. Then the value of the integral $\int_0^1 f(x) g(x) d x$, is
$\mathrm{e}+\frac{\mathrm{e}^2}{2}+\frac{5}{2}$
$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{5}{2}$
$\mathrm{e}+\frac{\mathrm{e}^2}{2}-\frac{3}{2}$
$\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$
Solution
Let $f(x)=e^x$
$\therefore \int_0^1 f(x) g(x) d x=\int_0^1 e^x\left(x^2-e^x\right) d x$
$=\int_0^1 x^2 e^x d x-\int_0^1 e^{2 x} d x$
$=\left[\mathrm{x}^2 \mathrm{e}^{\mathrm{x}}\right]_0^1-2\left[\mathrm{xe}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}\right]_0^1-\frac{1}{2}\left[\mathrm{e}^{2 \mathrm{x}}\right]$
$=\mathrm{e}-\left[\frac{\mathrm{e}^2}{2}-\frac{1}{2}\right]-2[\mathrm{e}-\mathrm{e}+1]=\mathrm{e}-\frac{\mathrm{e}^2}{2}-\frac{3}{2}$