Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If

$6 \int\limits_1^x f(t) d t=3 x f(x)-x^3$

for all $x \geq 1$, then the value of $f(2)$ is

6

We have,

$6 \int\limits_1^x f(t) d t=3 x f(x)-x^3$

Differentiating both sides with respect to x, we get

$6 f(x)=3 f(x)+3 x f'(x)-3 x^2$

$\Rightarrow 3 f(x)=3 x f'(x)-3 x^2$

$\Rightarrow f'(x)-\frac{1}{x} f(x)=x$

$\Rightarrow \frac{d}{d x}(f(x))+\left(-\frac{1}{x}\right) f(x)=x$

This is a linear differential equation with integrating factor

$e^{\int-\frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}$

Solution is given by

$\frac{1}{x} f(x)=\int x \times \frac{1}{x} d x+C $

$\Rightarrow f(x)=x^2+C x$

∴  $f(1)=1+C \Rightarrow 2=1+C \Rightarrow C=1$

Thus, $f(x)=x^2+x$

∴  $f(2)=4+2=6$