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(A) Given, $f(x) = \int_{0}^{x} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign, we differentiate both sides with r

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(A) Given, $f(x) = \int_{0}^{x} f(t) \, dt$. Applying the Leibniz rule for differentiation under the integral sign, we differentiate both sides with respect to $x$: $f'(x) = f(x)$. This is a first-order linear differential equation. Separating variables, we get $\frac{f'(x)}{f(x)} = 1$. Integrating both sides with respect to $x$, we get $\ln|f(x)| = x + C$, which implies $f(x) = k e^{x}$ for some constant $k$. Now, substitute $x = 0$ into the original equation: $f(0) = \int_{0}^{0} f(t) \, dt = 0$. Using $f(0) = k e^{0} = k$, we find $k = 0$. Therefore, $f(x) = 0 \cdot e^{x} = 0$ for all $x \in R$. Thus, $f(\log_{e} 5) = 0$.

Let $f: R \rightarrow R$ be a continuous function which satisfies $f(x) = \int_{0}^{x} f(t) \, dt$. Then, the value of $f(\log_{e} 5)$ is:

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