Function and Calculus.

$f(x) = 2 + \int_{2}^{x}(t - tf(t))dt$

Differentiating both sides w.r.t $x$ , we get

$f'(x) = x - xf(x)$

This is a linear differential equation. Its solution is $(e^{(\frac{x^{2}}{2})})f(x) =\int xe^{(\frac{x^{2}}{2})} + c$

$\implies$ $(e^{(\frac{x^{2}}{2})})f(x) =e^{(\frac{x^{2}}{2})} + c$

Since $f(2) = 2$ , so $c = e^2$ .

So, $f(x) = 1 + (e^2)(e^{-(\frac{x^{2}}{2})})$

Hence $f(0) = 1 + e^2 = 8.389$