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(C) Given the equation: $3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}$.Differentiating both sides with respect to $x$ using the Leibniz rule:$3 f(x) =

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$\frac{4e^2}{3}$

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(C) Given the equation: $3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}$.Differentiating both sides with respect to $x$ using the Leibniz rule:$3 f(x) = f(x) + x f'(x) - x^2$.Rearranging the terms:$2 f(x) = x f'(x) - x^2 \implies x f'(x) - 2 f(x) = x^2$.Dividing by $x$ $(x \geq 1)$:$f'(x) - \frac{2}{x} f(x) = x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = x$.Integrating Factor $(IF)$ = $e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2} = \frac{1}{x^2}$.The solution is given by $f(x) \cdot IF = \int Q(x) \cdot IF dx + C$:$f(x) \cdot \frac{1}{x^2} = \int x \cdot \frac{1}{x^2} dx = \int \frac{1}{x} dx = \ln x + C$.So,$f(x) = x^2 \ln x + C x^2$.Using the condition $f(1) = \frac{1}{3}$:$f(1) = 1^2 \ln(1) + C(1)^2 = \frac{1}{3} \implies 0 + C = \frac{1}{3} \implies C = \frac{1}{3}$.Thus,$f(x) = x^2 \ln x + \frac{x^2}{3}$.Calculating $f(e)$:$f(e) = e^2 \ln(e) + \frac{e^2}{3} = e^2(1) + \frac{e^2}{3} = \frac{3e^2 + e^2}{3} = \frac{4e^2}{3}$.

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

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