Question

Let $f\left(x\right)$ be a function satisfying $f^{\prime }\left(x\right)=f\left(x\right)=e^x$ 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 $e-\frac{e^2}{2}-\frac{3}{2}$

Let $f^{\prime }\left(x\right)=f\left(x\right)=k{e}^x$
As $f\left(0\right)=1\Rightarrow k=1$
Gives $f\left(x\right)=e^x\Rightarrow g\left(x\right)=x^2-e^x$
Therefore,
${\int }_0^{1}f\left(x\right)g\left(x\right)dx={\int }_0^1{e}^x(x^2-e^x)dx={\int }_0^1{e}^x{x}^{2}dx-{\int }_0^1{e}^{2x}dx={[e^x{x}^2-2x{e}^x+2e^x-\frac{e^{2x}}{2}]}_0^1=e-\frac{e^2}{2}-\frac{3}{2}$