Mathematics - Definite Integration Question with Solution | TestHub

$f(x)=e^x{\int }_0^1{e}^{t}f(t)dt$

$f(x)=A{e}^x$

$\Rightarrow A={\int }_0^1{e}^{t}A{e}^{t}dt\Rightarrow A=0$

$\Rightarrow f(x)=0$

$g(x)=e^x{\int }_0^1{e}^{t}g(t)dt+x$

$g(x)=B{e}^x+x$

$B={\int }_0^1{e}^t\left(B{e}^t+t\right)dt$

$B=B{\left[\frac{e^{2t}}{2}\right]}_0^1+{\int }_0^{1}t{e}^{t}dt$

$$\begin{array}{rl}& B=\frac{B}{2}\left[e^2-1\right]+{\left[t{e}^t+e^t\right]}_0^1\\ & B\left[1-\frac{e^2}{2}+\frac{1}{2}\right]=1+1\\ & B=\frac{-2}{e^2-3}\\ & g(0)=\frac{-2}{e^2-3}\Rightarrow g(0)=\frac{2}{3-e^2}\\ & g(x)=\left(\frac{2}{3-e^2}{e}^x+x\right)\\ & g(2)=2\left[\frac{e^2}{3-e^2}+1\right]\Rightarrow g(2)=2\left(\frac{3}{3-e^2}\right)=\left(\frac{6}{3-e^2}\right)\\ & \frac{g(0)}{g(2)}=\frac{2}{6}=\frac{1}{3}\end{array}$$