Let f:[1, ∞) -> [2, ∞) be a differentiable function such that f(1) = 2. If 6∫[x, 1] f(t) dt = 3xf(x) - x³ for all x ≥ 1, then the value of f(2) is :

6∫[x, 1] f(t) dt = 3xf(x) - x³
Differentiating w.r.t. x, we get (Use Newton Leibnitz theorem for differentiating a definite integral)-> 6f(x) = 3xf'(x) + 3f(x) - 3x²
-> xf'(x) = f(x) + x²
or, dy/dx - y/x = x
This is a first order linear differential equation with Integrating factor e^(-∫1/x dx) = e^(-lnx) = 1/x
Its general solution is y/x = ∫x * 1/x dx + C
-> y/x = x + C
Since f(1) = 2
-> C = 1
So, y = x² + x
-> f(2) = 6