Definite Integration
Limit as Definite Integral
Grade 12
Question:
<p>The value of \(\lim_{n \to \infty}\left(\ln\left(\sqrt[n]{\dfrac{4}{n^2}}\right) + \ln\left(\sqrt[n]{\dfrac{16}{n^2}}\right) + \ln\left(\sqrt[n]{\dfrac{36}{n^2}}\right) + \ldots + \ln\left(\sqrt[n]{\dfrac{4n^2}{n^2}}\right)\right)\) equals:</p>
<p>(a) \(4\ln(2)\)</p>
<p>(b) \(2\ln(2) - 2\)</p>
<p>(c) \(2\ln(2) - 4\ln(4) - 4\)</p>
<p>(d) \(2\ln(4) - 2\)</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum by converting the logarithm sum into an integral form. Each term ln(∜[n]{4k²/n²}) = (1/n)ln(2k/n), creating a partition of [0,2] with width 1/n.
<p><strong>Step 1:</strong> Simplify the general term</p><p>ln(∜[n]{4k²/n²}) = (1/n)·ln(4k²/n²) = (1/n)[ln(4k²) - ln(n²)] = (1/n)[ln(4) + 2ln(k) - 2ln(n)]</p><p><strong>Step 2:</strong> Recognize the sum structure as a Riemann sum</p><p>S = ∑(k=1 to n) (1/n)ln(4k²/n²) = (1/n)∑(k=1 to n) ln((2k/n)²) = (1/n)∑(k=1 to n) 2ln(2k/n)</p><p>= 2·(1/n)∑(k=1 to n) ln(2k/n)</p><p><strong>Step 3:</strong> Convert to integral form</p><p>Let f(x) = ln(2x). The sum is a Riemann sum for ∫₀² ln(2x)dx with partition width Δx = 1/n</p><p>lim(n→∞) = 2∫₀² ln(2x)dx</p><p><strong>Step 4:</strong> Evaluate the integral</p><p>∫ln(2x)dx = x·ln(2x) - x + C</p><p>∫₀² ln(2x)dx = [x·ln(2x) - x]₀² = (2ln4 - 2) - lim(x→0⁺)(x·ln(2x) - x) = 2ln4 - 2 - 0 = 2ln4 - 2</p><p><strong>Step 5:</strong> Compute final answer</p><p>2∫₀² ln(2x)dx = 2(2ln4 - 2) = 4ln4 - 4 = 4ln(2²) - 4 = 8ln2 - 4</p><p>∴ Answer: <strong>8ln2 - 4 or 4(2ln2 - 1)</strong></p>
Correct Answer: D