Question

Let f(x) be a function satisfying f ’(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 ${\int }_0^{1}f\left(x\right)g\left(x\right)dx,$ is

• A. $e-\frac{e^2}{2}-\frac{5}{2}$
• B. $e+\frac{e^2}{2}-\frac{3}{2}$
• C. $e-\frac{e^2}{2}-\frac{3}{2}$
• D. $e+\frac{e^2}{2}+\frac{5}{2}$

Answer 1

The correct option is C $e-\frac{e^2}{2}-\frac{3}{2}$

Given f’(x) = f(x) and f(0) = 1
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=1\Rightarrow lnf\left(x\right)=x+c-------\left(1\right)$
f(0)=1
$\Rightarrow c=0$
$\Rightarrow Inf\left(x\right)=x\Rightarrow f\left(x\right)=e^x$
$\Rightarrow g\left(x\right)=x^2-e^x$
$\therefore f\left(x\right)g\left(x\right)=e^x(x^2-e^x)={\int }_0^1(e^x{x}^2-e^{2x})dx$
$={\left[\right(x^2-2x+2\left)e^x\right]}_0^1-\frac{1}{2}{e}^2+\frac{1}{2}$
$=e-\frac{1}{2}{e}^2-\frac{3}{2}$