Let $f : [1, \infty) \to \mathbb{R}$ be a differentiable function such that $f(1) = \frac{1}{3}$ and $3\int_{1}^{x} f(t)\,dt = x f(x) - \frac{x^{3}}{3}$, $x \in [1, \infty)$. Let $e$ denote the base of the natural logarithm. Then the value of $f(e)$ is

Let f : [1, ∞) → R be a differentiable function such that f(1) = 1/3 and 3∫₁ˣ f(t) dt = x f(x) - x³/3, x ∈ [1, ∞). Let e denote the base of the natural logarithm. Then the value of f(e) is $f(1) = \dfrac{1}{3}$ $3\int_{1}^{x} f(t)\,dt = x f(x) - \dfrac{x^3}{3}, \quad x\in[1,\infty)$ $\text{To find: } f(e)$ $3 f(x) = f(x) + x f'(x) - x^2 \Rightarrow x f'(x) = 2 f(x) + x^2$ $f'(x) - \dfrac{2}{x}f(x) = x$ $\text{Integrating factor} = e^{\int -\frac{2}{x}dx} = e^{-2\ln x} = x^{-2}$ $(f x^{-2})' = x\cdot x^{-2} = \dfrac{1}{x}$ $f x^{-2} = \ln x + C \Rightarrow f(x) = x^2(\ln x + C)$ $f(1) = 1^2(0 + C) = \dfrac{1}{3} \Rightarrow C = \dfrac{1}{3}$ $f(x) = x^2\left(\ln x + \dfrac{1}{3}\right) \Rightarrow f(e) = e^2\left(1+\dfrac{1}{3}\right) = \dfrac{4e^2}{3}$ $\boxed{f(e) = \dfrac{4e^2}{3}} \quad \text{option (3)}$