Let f:[1, ∞) arrow[2, ∞) be a differentiable function such that...

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 $\quad f(2)=4-\frac{4}{3}=\frac{8}{3}$