Question

Let f(x) be a function satisfying $f^{{}^{\prime }}\left(x\right)=f\left(x\right)$ with f(0) = 1
and g(x) be the function satisfying $f\left(x\right)+g\left(x\right)=x^2$
The value of integral ${\int }_0^{1}f\left(x\right)g\left(x\right)dx$ is equal to [AIEEE 2003; DCE 2005]

• A. $\frac{1}{4}(e-7)$
• B. $\frac{1}{4}(e-2)$
• C. $\frac{1}{2}(e-3)$
• D. $Noneofthese$

Answer 1

The correct option is D

$Noneofthese$

We have $f^{{}^{\prime }}\left(x\right)=f\left(x\right)\Rightarrow \frac{f^{{}^{\prime }}\left(x\right)}{f\left(x\right)}=1$
$\Rightarrow logf\left(x\right)=x+logc\Rightarrow f\left(x\right)=c{e}^x$
Since f(0)=1, therefore $1=c{e}^0\Rightarrow c=1$
Thus $f\left(x\right)=e^x$
Hence $g\left(x\right)=x^2-e^x$
$\therefore {\int }_0^{1}f\left(x\right)g\left(x\right)dx={\int }_0^1{e}^x(x^2-e^x)dx$
$=e-\frac{1}{2}{e}^2-\frac{3}{2}$