Question

Let $f\left(x\right)$ be a function satisfies $f^{\prime }\left(x\right)=f\left(x\right)$ with $f\left(0\right)=1$ and $g\left(x\right)$ be a function that satisfies $f\left(x\right)+g\left(x\right)=x^2$, then the value of the integral ${\int }_0^{1}f\left(x\right)g\left(x\right)dx$ is

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

Answer 1

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

As $f\left(x\right)=f^{\prime }\left(x\right)$ and $f\left(0\right)=1$
$\Rightarrow$ $\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=1$
$\Rightarrow$ $\log \left(f\left(x\right)\right)=x$
$\Rightarrow$ $f\left(x\right)=e^x+k$
$\Rightarrow$ $f\left(x\right)=e^x$ as $f\left(0\right)=1$
Now $g\left(x\right)=x^2-e^x$
$\therefore$ ${\int }_0^{1}f\left(x\right)g\left(x\right)dx={\int }_0^1{e}^x(x^2-e^x)dx$
$={\int }_0^1{x}^2{e}^{x}dx-{\int }_0^1{e}^{2x}dx$
$={\left[(x^2-2x+2)e^x\right]}_0^1-{\left(\frac{e^{2x}}{2}\right)}_0^1$
$=(e-2)-\left(\frac{e^2-1}{2}\right)=e-\frac{e^2}{2}-\frac{3}{2}$
Using $f^{n}x{e}^{x}dx=e^x[f^n\left(x\right)-f_1^n\left(x\right)+f_2^n\left(x\right)+...+{(-1)}^n{f}_n\left(x\right)]$
where $f_1$, $f_2$, ..., $f_n$ are derivatives of first, second , ...n${}^{th}$ order.