Question

If $f\left(x\right)$ is a function satisfying $f^{\prime }\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and $g\left(x\right)$ be another function such that $f\left(x\right)+g\left(x\right)=x^2$, then the value of $\underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx$ is

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

Answer 1

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

$f^{\prime }\left(x\right)=f\left(x\right)\Rightarrow \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\int dx+C\Rightarrow \ln |f\left(x\right)|=x+C\Rightarrow f\left(x\right)=a{e}^x$

Given $f\left(0\right)=1$, we get
$f\left(x\right)=e^x$
We know that,
$f\left(x\right)+g\left(x\right)=x^2\Rightarrow g\left(x\right)=x^2-e^x$

Now,
$\underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx=\underset{0}{\overset{1}{\int }}{e}^x(x^2-e^x)dx=\underset{0}{\overset{1}{\int }}{x}^2{e}^{x}dx-\underset{0}{\overset{1}{\int }}{e}^{2x}dx={[x^2{e}^x-2x{e}^x+2e^x]}_0^1-\frac{1}{2}{\left[e^{2x}\right]}_0^1=[e-2]-\frac{1}{2}[e^2-1]=e-2-\frac{e^2}{2}+\frac{1}{2}=e-\frac{e^2}{2}-\frac{3}{2}$

Hence,
$\underset{0}{\overset{1}{\int }}f\left(x\right)g\left(x\right)dx=\frac{1}{2}(2e-e^2-3)$