Definite Integration Question - None

Question

35. 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: Use the given functional equation: xf(x) = x + ∫₁ˣ f(t)dt

Step 2: Differentiate both sides with respect to x:

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

This simplifies to: xf'(x) = 1, so f'(x) = 1/x

Step 3: Integrate to get: f(x) = ln(x) + C

Step 4: Apply initial condition. Substitute x = 1 in original equation:

1·f(1) = 1 + ∫₁¹ f(t)dt = 1 + 0

Therefore f(1) = 1, which gives: ln(1) + C = 1, so C = 1

Step 5: Thus f(x) = ln(x) + 1

Step 6: Calculate the sum:

∑ₖ₌₁¹⁰ f(eᵏ) = ∑ₖ₌₁¹⁰ [ln(eᵏ) + 1] = ∑ₖ₌₁¹⁰ (k + 1)

= ∑ₖ₌₁¹⁰ (k + 1) = 2 + 3 + 4 + ... + 11 = (1 + 2 + ... + 11) - 1 = 66 - 1 = 65

∴ Answer: A (65)

Question:

Step-by-Step Solution

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