What is 8\int_{1}^{2}f(x)\,dx equal to ?
Consider the following for the next two (02) items that follow : Let 3f(x)+f(\frac{1}{x})=\frac{1}{x}+1
- A. \ln(8\sqrt{e}) ✓
- B. \ln(4\sqrt{e})
- C. \ln 2
- D. \ln 2-1
Correct Answer: A. \ln(8\sqrt{e})
Explanation
From the previous solution, 8f(x) = \frac{3}{x} - x + 2. Integrating this expression from 1 to 2: \int_1^2 (\frac{3}{x} - x + 2)\,dx = \left[ 3\ln x - \frac{x^2}{2} + 2x \right]_1^2. Evaluating at upper limit 2: 3\ln 2 - 2 + 4 = 3\ln 2 + 2. Evaluating at lower limit 1: 3\ln 1 - \frac{1}{2} + 2 = 0 + \frac{3}{2}. Subtracting the lower limit value gives 3\ln 2 + 2 - \frac{3}{2} = 3\ln 2 + \frac{1}{2}. This can be rewritten as \ln(2^3) + \ln(e^{1/2}) = \ln 8 + \ln(\sqrt{e}) = \ln(8\sqrt{e}).
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