Let  $f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt$  and  $f\left(0\right)=0.$ Then  $\left[f\left(3\right)\right]$  where [.] represent greatest integer

$f\left(x\right)=x^2+\underset{0}{\overset{x}{\int }}{e}^{-t}f(x-t)dt=x^2+\underset{0}{\overset{x}{\int }}{e}^{-(x-t)}f\left(t\right)dt=x^2+\underset{0}{\overset{x}{\int }}{e}^{-x}.e^{t}f\left(t\right)dt=x^2+e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt$$f^1\left(x\right)=2x+e^{-x}.e^{x}f\left(x\right)-e^{-x}\underset{0}{\overset{x}{\int }}{e}^{t}f\left(t\right)dt=2x+f\left(x\right)-f\left(x\right)+x^2$$f\left(x\right)=x^2+\frac{x^3}{3}+c$

$f\left(0\right)=0$

$f\left(x\right)=x^2+\frac{x^3}{3}$

$f\left(3\right)=9+9=18$