Question:

Step 1: Differentiate the given functional equation to find a differential equation. We are given that $xf(x) = x + \displaystyle\int_{1}^{x} f(t)\, dt$. Differentiating both sides with respect to $x$ using the product rule on the left side and the fundamental theorem of calculus on the right side: $$\frac{d}{dx}[xf(x)] = \frac{d}{dx}\left[x + \displaystyle\int_{1}^{x} f(t)\, dt\right]$$ $$f(x) + xf'(x) = 1 + f(x)$$ Step 2: Simplify the differential equation. Subtracting $f(x)$ from both sides: $$xf'(x) = 1$$ $$f'(x) = \frac{1}{x}$$ Step 3: Integrate to find the general form of $f(x)$. Integrating both sides with respect to $x$: $$f(x) = \displaystyle\int \frac{1}{x}\, dx = \ln|x| + C$$ Since $f(x)$ is defined for all positive real numbers, we can write: $$f(x) = \ln x + C$$ Step 4: Find the constant of integration using the original equation. Substitute $f(x) = \ln x + C$ into the original equation $xf(x) = x + \displaystyle\int_{1}^{x} f(t)\, dt$: $$x(\ln x + C) = x + \displaystyle\int_{1}^{x} (\ln t + C)\, dt$$ Evaluate the integral on the right side: $$\displaystyle\int_{1}^{x} (\ln t + C)\, dt = \displaystyle\int_{1}^{x} \ln t\, dt + \displaystyle\int_{1}^{x} C\, dt$$ Using integration by parts for $\displaystyle\int \ln t\, dt = t\ln t - t$: $$= [t\ln t - t]_{1}^{x} + C[t]_{1}^{x} = (x\ln x - x) - (0 - 1) + C(x - 1)$$ $$= x\ln x - x + 1 + Cx - C$$ Step 5: Simplify to find the value of $C$. Substituting back into the equation: $$x\ln x + Cx = x + x\ln x - x + 1 + Cx - C$$ $$x\ln x + Cx = x\ln x + Cx + 1 - C$$ $$0 = 1 - C$$ $$C = 0$$ Therefore: $f(x) = \ln x$ Step 6: Calculate the sum $\displaystyle\sum_{k=1}^{10} f(e^k)$. We need to find: $$\displaystyle\sum_{k=1}^{10} f(e^k) = \displaystyle\sum_{k=1}^{10} \ln(e^k) = \displaystyle\sum_{k=1}^{10} k$$ Step 7: Evaluate the arithmetic series. The sum $\displaystyle\sum_{k=1}^{10} k$ is an arithmetic series with first term $a_1 = 1$, last term $a_{10} = 10$, and $n = 10$ terms. Using the arithmetic series formula: $$\displaystyle\sum_{k=1}^{10} k = \frac{n(a_1 + a_n)}{2} = \frac{10(1 + 10)}{2} = \frac{10 \times 11}{2} = 55$$ Therefore, the value of $\displaystyle\sum_{k=1}^{10} f(e^k)$ is $\boxed{55}$, which corresponds to **Option 2**.