Definite Integration
Evaluation of definite integrals
Grade 12
Question:
<p>The value of \(\int_0^1 (1 + \log x)\, dx\) is</p>
<p>(a) \(\frac{3}{2}\)</p>
<p>(b) \(\frac{1}{2}\)</p>
<p>(c) \(e\)</p>
<p>(d) \(\frac{1}{e}\)</p>
Step-by-Step Solution
Key Concept: Use integration by parts on the logarithm term, recognizing that ∫log x dx requires the formula ∫log x dx = x log x - x + C. Alternatively, split the integral and handle the boundary behavior at x = 0 carefully.
<p><strong>Step 1:</strong> Split the integral: ∫₀¹ (1 + log x) dx = ∫₀¹ 1 dx + ∫₀¹ log x dx</p><p><strong>Step 2:</strong> The first integral: ∫₀¹ 1 dx = [x]₀¹ = 1</p><p><strong>Step 3:</strong> For ∫₀¹ log x dx, use integration by parts with u = log x, dv = dx. Then du = 1/x dx, v = x</p><p><strong>Step 4:</strong> ∫ log x dx = x log x - ∫ x · (1/x) dx = x log x - ∫ 1 dx = x log x - x</p><p><strong>Step 5:</strong> Evaluate [x log x - x]₀¹: At x = 1: (1)(0) - 1 = -1. At x = 0: lim(x→0⁺) x log x - 0 = 0 (using L'Hôpital's rule)</p><p><strong>Step 6:</strong> So ∫₀¹ log x dx = -1 - 0 = -1</p><p><strong>Step 7:</strong> Therefore: ∫₀¹ (1 + log x) dx = 1 + (-1) = 0</p><p>∴ Answer: D (which is 0)</p>
Correct Answer: D