Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>For <span>\(x > 0\)</span>, let <span>\(f(x) = \int_{1}^{x} \frac{\log t}{1+t} dt\)</span>. Then <span>\(f(x) + f\left(\frac{1}{x}\right)\)</span> is equal to:</p>
<p>\(\frac{1}{4}(\log x)^2\)</p>
<p>\(\frac{1}{2}(\log x)^2\)</p>
<p>\(\log x\)</p>
<p>\((\log x)^2\)</p>
Step-by-Step Solution
Key Concept: Use substitution u = 1/t in f(1/x) to relate it to f(x), then exploit the symmetric property that f(x) + f(1/x) involves integrals over complementary domains that can be combined using a clever algebraic manipulation.
<p><strong>Step 1:</strong> Write f(1/x) using substitution. Let t = 1/u in the integral definition:</p><p>$$f\left(\frac{1}{x}\right) = \int_{1}^{1/x} \frac{\log t}{1+t} dt$$</p><p>Substitute u = 1/t, so t = 1/u, dt = -du/u². When t = 1, u = 1; when t = 1/x, u = x:</p><p>$$f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log(1/u)}{1+(1/u)} \cdot \frac{-du}{u^2} = \int_{1}^{x} \frac{-\log u}{1+u} du$$</p><p><strong>Step 2:</strong> Add f(x) and f(1/x):</p><p>$$f(x) + f\left(\frac{1}{x}\right) = \int_{1}^{x} \frac{\log t}{1+t} dt + \int_{1}^{x} \frac{-\log t}{1+t} dt$$</p><p>$$= \int_{1}^{x} \frac{\log t - \log t}{1+t} dt = \int_{1}^{x} \frac{0}{1+t} dt = 0$$</p><p><strong>Step 3:</strong> Verify: The integrals cancel exactly because they are negatives of each other over the same domain.</p><p>∴ Answer: <strong>f(x) + f(1/x) = 0</strong></p>
Correct Answer: B