Question

Let $f^{\prime }\left(x\right)=f\left(x\right)$ where $f\left(0\right)=1.$ If $f\left(x\right)+g\left(x\right)=x^2$ then ${\int }_0^{1}f\left(x\right)g\left(x\right)dx$ is equal to:

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

Answer 1

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

Given that $f^{\prime }\left(x\right)=f\left(x\right)\Rightarrow f\left(x\right)=c{e}^x$

And since $f\left(0\right)=1$

$\therefore 1=f\left(0\right)=c\Rightarrow f\left(x\right)=e^x$

Hence $g\left(x\right)=x^2-e^x$

Thus ${\int }_0^{1}f\left(x\right)g\left(x\right)dx={\int }_0^1{e}^x(x^2-e^x)dx$

$={\left[x^2{e}^x\right]}_0^1-2{\int }_0^{1}x{e}^{x}dx-{\left[\frac{e^{2x}}{2}\right]}_0^1$

$=(e-0)-2{\left[x{e}^x\right]}_0^1-{\left[e^x\right]}_0^1-\frac{1}{2}(e^2-1)$

$=(e-0)-2[(e-0)-(e-1)]-\frac{1}{2}(e^2-1)$

$=e-\frac{1}{2}{e}^2-\frac{3}{2}$