Let $f\left(x\right)$ be a polynomial, with positive leading coefficient, satisfying $f\left(0\right)=0$ and $\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\forall \mathrm{x}\in \mathrm{R}.$ Then $f\left(2\right)$ is equal to

f(x) is polynomial of define n, then f(f(x)) is polynomial of degree 2n and $\mathrm{x}\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$ is polynomial of degree n + 2
As $\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\_\_\_\_\_\left(1\right)$
2n = n + 2
n = 2
f(x) is polynomial of degree 2
$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}\mathrm{as}\mathrm{f}\left(0\right)=0$
Differential (1) w.r.t. x
$\mathrm{f}\text{'}\left(\mathrm{f}\left(\mathrm{x}\right)\right).\mathrm{f}\text{'}\left(\mathrm{x}\right)=\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}+\mathrm{xf}\left(\mathrm{x}\right)$
$\mathrm{f}\text{'}\left(0\right)=0$
Hence, $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2$
Now, $\mathrm{f}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\mathrm{x}\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$
$\mathrm{a}{\left({\mathrm{ax}}^2\right)}^2=\mathrm{x}\underset{0}{\overset{\mathrm{x}}{\int }}{\mathrm{at}}^2\mathrm{dt}=\frac{{\mathrm{ax}}^4}{3}$
${\mathrm{a}}^3=\frac{\mathrm{a}}{3}\Rightarrow \mathrm{a}=0,\pm \frac{1}{\sqrt{3}}$
$\mathrm{f}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2}{\sqrt{3}}$