Question

Let f (x) be function satisfying $f^{\prime }\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and g be the function satisfying f(x)+g(x)=$x^2.$ The value of the integral ${\int }_0^{1}f\left(x\right)g\left(x\right)dx$ is

• A. $e-\frac{1}{2}{e}^2-\frac{5}{2}$
• B. $e-e^2-3$
• C. $\frac{1}{2}(e-3)$
• D. $e-\frac{1}{2}{e}^2-\frac{3}{2}$

Answer 1

The correct option is D $e-\frac{1}{2}{e}^2-\frac{3}{2}$

$f^1\left(x\right)=f\left(x\right)\Rightarrow f\left(x\right)=c{e}^x$
$\because f\left(0\right)=1\therefore 1=f\left(0\right)=c\therefore f\left(x\right)=e^x\Rightarrow g\left(x\right)=x^2-e^x={\int }_0^1(x^2{e}^x-e^{2x})dx$
$={\left[x^2{e}^x\right]}_0^1-2{\int }_0^1{xe}^{x}dx-{\left[\frac{e^{2x}}{2}\right]}_0^1$
$=(e-0)-2\left({\left[{xe}^x\right]}_0^1-{\left[e^x\right]}_0^1\right)-\frac{1}{2}(e^2-1)$
$=(e-0)-2[(e-0)-(e-1)]-\frac{1}{2}(e^2-1)$
$=e-\frac{e^2}{2}-\frac{3}{2.}$