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 $f\left(x\right)+g\left(x\right)=x$, then the value of $\stackrel{1}{\underset{0}{\int }}f\left(x\right)g\left(x\right)dx$

• A. $\frac{3-e^2}{2}$
• B. $\frac{e^2-3}{2}$
• C. $\frac{e^2}{2}$
• D. $\frac{e-2}{4}$

Answer 1

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

Given, $f^{\prime }\left(x\right)=f\left(x\right)$

Integrating $\log f\left(x\right)=x+k$
$\Rightarrow f\left(x\right)=e^{x+k}$

$=>f\left(0\right)=1$

$=>1=e^0.e^k$

$=>k=0$

$\Rightarrow f\left(x\right)=e^x$
$\Rightarrow g\left(x\right)=x-f\left(x\right)=x-e^x$

$I=\stackrel{1}{\underset{0}{\int }}{e}^x\left(\begin{array}{c}x-e^x\end{array}\right)dx=\stackrel{1}{\underset{0}{\int }}{e}^{x}xdx-\stackrel{1}{\underset{0}{\int }}{e}^{2x}dx$

$I=x\stackrel{1}{\underset{0}{\int }}{e}^{x}dx-\stackrel{1}{\underset{0}{\int }}1\cdot e^{x}dx-\stackrel{1}{\underset{0}{\int }}{e}^{2x}dx$

$I={[x{e}^x-e^x-\frac{e^{2x}}{2}]}_0^1$

$=e-(e-1)-\frac{e^2}{2}+\frac{1}{2}=\frac{3}{2}-\frac{e^2}{2}=\frac{3-e^2}{2}$.