Let \(f:[1, \infty) \rightarrow[2, \infty)\) be a differentiable function such that \(f(1)=2\). If \(6 \int_1^x f(t) d t=3 x f(x)-x^3\) for all \(x \geq 1\), then the value of \(f(2)\) is

Given, \(f(1)=\frac{1}{3}\) and \(6 \int_1^x f(t) d t\) \(=3 x f(x)-x^3\), for all \(x \geq 1\)
Using (Newton-Leibnitz formula),
On Differentiating both sides,
\[
\begin{aligned}
& 6 f(x) \cdot 1-0-3 f(x)+3 x f^{\prime}(x)-3 x^2 \\
\Rightarrow & 3 x f^{\prime}(x)-3 f(x)=3 x^2 \\
\Rightarrow & f^{\prime}(x)-\frac{1}{x} f(x)=x
\end{aligned}
\]

\[
\Rightarrow \frac{x f^{\prime}(x)-f(x)}{x^2}=1 \Rightarrow \frac{d}{d x}\left\{\frac{f(x)}{x}\right\}=1
\]
On integrating both sides,
\[
\begin{array}{ll}
& \frac{f(x)}{x}=x+C \quad\left[\because f(1)=\frac{1}{3}\right] \\
\Rightarrow \quad & \frac{1}{3}=1+C \Rightarrow \quad C=-\frac{2}{3} \\
& \text { Now, } \quad f(x)=x^2-\frac{2}{3} x \\
\Rightarrow \quad & f(2)=4-\frac{4}{3}=\frac{8}{3}
\end{array}
\]
Note Here, \(f(1)=2\) does not satisfy given function.
\[
\therefore \quad f(1)=\frac{1}{3}
\]
For that, \(f(x)=x^2-\frac{2}{3} x\) and \(\quad f(2)=4-\frac{4}{3}=\frac{8}{3}\)