Question

Let $f\left(x\right)$ be a function satisfying $f^{\prime }\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and $g$ be the function satisfying $f\left(x\right)+g\left(x\right)=x^2$. 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. $e+\frac{e^2}{2}$

Answer 1

Answer: (A), (D)

The correct options are
A $e-\frac{e^2}{2}-\frac{5}{2}$
D $e+\frac{e^2}{2}$

$f^{\prime }\left(x\right)=f\left(x\right)\Rightarrow f\left(x\right)=c{e}^x$
And since $f\left(0\right)=1$
$\therefore 1=f\left(0\right)=c\Rightarrow f\left(x\right)=e^x$
Hence $g\left(x\right)=x^2-e^x$
Thus ${\int }_0^{1}f\left(x\right)g\left(x\right)dx={\int }_0^1{e}^x(x^2-e^x)dx$
$={\left[x^2{e}^x\right]}_0^1-2{\int }_0^{1}x{e}^{x}dx-{\left[\frac{e^{2x}}{2}\right]}_0^1$
$=(e-0)-2({\left[x{e}^x\right]}_0^1-{\left[e^x\right]}_0^1)-\frac{1}{2}(e^2-1)$
$=(e-0)-2[(e-0)-(e-1)]-\frac{1}{2}(e^2-1)$
$=e-\frac{1}{2}{e}^2-\frac{3}{2}$