Definite Integration Question - None

Question

If $f(x)$ is a differentiable function defined for all positive real numbers such that $xf(x) = x + \int_{1}^{x} f(t)\, dt$, then the value of $\sum_{k=1}^{10} f(e^k)$ is:

Accepted Answer

Step 1: Given: $xf(x) = x + \int_{1}^{x} f(t)\, dt$

Step 2: Differentiate both sides with respect to $x$:

$\frac{d}{dx}[xf(x)] = \frac{d}{dx}\left[x + \int_{1}^{x} f(t)\, dt\right]$

$f(x) + xf'(x) = 1 + f(x)$

Step 3: Simplify: $xf'(x) = 1$, so $f'(x) = \frac{1}{x}$

Step 4: Integrate: $f(x) = \ln x + C$

Step 5: Use initial condition. Set $x=1$ in original equation: $1 \cdot f(1) = 1 + 0$, so $f(1) = 1$

Therefore: $\ln 1 + C = 1 \Rightarrow C = 1$

Thus: $f(x) = \ln x + 1$

Step 6: Calculate the sum:

$\sum_{k=1}^{10} f(e^k) = \sum_{k=1}^{10} (\ln e^k + 1) = \sum_{k=1}^{10} (k + 1) = \sum_{k=1}^{10} k + 10$

$= \frac{10 \cdot 11}{2} + 10 = 55 + 10 = 65$

∴ Answer: B (65)

Question:

Step-by-Step Solution

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