Evaluation of Function Value Using Integral Equation and Differentiation

To find the value of $f(e)$, we begin by differentiating the given integral equation with respect to $x$ using Leibniz's rule.

The given integral equation is: $3 \int_{1}^{x} f(t) \, dt = x f(x) - \frac{x^3}{3}, \quad x \in [1, \infty)$

Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( 3 \int_{1}^{x} f(t) \, dt \right) = \frac{d}{dx} \left( x f(x) - \frac{x^3}{3} \right)$

Applying Leibniz's rule on the left side and the product rule on the right side, we get: $3 f(x) = f(x) + x f'(x) - x^2$

Rearranging the terms: $2 f(x) = x f'(x) - x^2$ $x f'(x) - 2 f(x) = x^2$

Since $x \ge 1$, we can divide the entire equation by $x$: $f'(x) - \frac{2}{x} f(x) = x$

This is a first-order linear differential equation of the form $\frac{df}{dx} + P(x)f = Q(x)$, where $P(x) = -\frac{2}{x}$ and $Q(x) = x$.

The integrating factor ($\text{I.F.}$) is given by: $\text{I.F.} = e^{\int -\frac{2}{x} \, dx} = e^{-2 \ln x} = e^{\ln(x^{-2})} = \frac{1}{x^2}$

Multiplying the differential equation by the integrating factor yields: $\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = x \cdot \frac{1}{x^2} = \frac{1}{x}$

Integrating both sides with respect to $x$: $\frac{f(x)}{x^2} = \int \frac{1}{x} \, dx = \ln x + C$ $f(x) = x^2 (\ln x + C)$

We are given the initial condition $f(1) = \frac{1}{3}$. Substituting $x = 1$ into the equation above: $f(1) = 1^2 \cdot (\ln 1 + C)$ $\frac{1}{3} = 0 + C \implies C = \frac{1}{3}$

Therefore, the function $f(x)$ is: $f(x) = x^2 \left( \ln x + \frac{1}{3} \right)$

Now, evaluating $f(e)$ by substituting $x = e$: $f(e) = e^2 \left( \ln e + \frac{1}{3} \right)$ $f(e) = e^2 \left( 1 + \frac{1}{3} \right) = \frac{4e^2}{3}$

Hence, the correct option is (C).