Let $$f: [1, \infty) \to \mathbb{R}$$ be a differentiable function such that $$f(1) = \frac{1}{3}$$ and $$3\int_1^x f(t)\,dt = xf(x) - \frac{x^3}{3}$$, $$x \in [1, \infty)$$. Let $$e$$ denote the base of the natural logarithm. Then the value of $$f(e)$$ is

The given relation is
$$3\int_{1}^{x} f(t)\,dt \;=\; x\,f(x)\;-\;\frac{x^{3}}{3},\qquad x\in[1,\infty).$$

Differentiating both sides with respect to $$x$$ (using the Fundamental Theorem of Calculus and the Product Rule):
Left side: $$\frac{d}{dx}\Bigl[3\int_{1}^{x} f(t)\,dt\Bigr]=3f(x).$$
Right side: $$\frac{d}{dx}\Bigl[x\,f(x)-\frac{x^{3}}{3}\Bigr]=f(x)+x\,f'(x)-x^{2}.$$

Equating the derivatives:
$$3f(x)=f(x)+x\,f'(x)-x^{2}.$$

Simplify:
$$2f(x)=x\,f'(x)-x^{2} \quad\Longrightarrow\quad x\,f'(x)-2f(x)=x^{2}.$$

Divide by $$x$$ to convert it into a linear first-order ODE:
$$f'(x)-\frac{2}{x}\,f(x)=x.$$

The integrating factor is
$$\mu(x)=\exp\!\Bigl(\int -\frac{2}{x}\,dx\Bigr)=\exp(-2\ln x)=x^{-2}.$$

Multiply the differential equation by $$\mu(x)=x^{-2}$$:
$$x^{-2}f'(x)-2x^{-3}f(x)=x^{-1}.$$

The left side is the derivative of $$x^{-2}f(x)$$:
$$\frac{d}{dx}\!\bigl[x^{-2}f(x)\bigr]=x^{-1}.$$

Integrate with respect to $$x$$:
$$x^{-2}f(x)=\int x^{-1}\,dx=\ln x + C.$$

Hence,
$$f(x)=x^{2}\,\bigl(\ln x + C\bigr).$$

Use the initial condition $$f(1)=\tfrac13$$:
$$\tfrac13 = 1^{2}\bigl(\ln 1 + C\bigr)=0 + C \;\Longrightarrow\; C=\tfrac13.$$

Therefore,
$$f(x)=x^{2}\Bigl(\ln x+\tfrac13\Bigr).$$

Substitute $$x=e$$ (where $$\ln e = 1$$):
$$f(e)=e^{2}\Bigl(1+\tfrac13\Bigr)=e^{2}\cdot\frac{4}{3}=\frac{4e^{2}}{3}.$$

Option C which is: $$\frac{4e^{2}}{3}$$