Definite Integration
Limit as sum
Grade 12
Question:
<p>\(\lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r}{\sqrt{n^2 + r^2}}\) equals</p>
<p>(a) \(1 + \sqrt{5}\)</p>
<p>(b) \(-1 + \sqrt{5}\)</p>
<p>(c) \(-1 + \sqrt{2}\)</p>
<p>(d) \(1 + \sqrt{2}\)</p>
Step-by-Step Solution
Key Concept: Recognize this limit as a Riemann sum by factoring out n from the denominator: rewrite as (1/n)∑f(r/n) where f(x) = x/√(1+x²), then convert to the definite integral ∫₀¹ x/√(1+x²) dx.
<p><strong>Step 1: Identify the Riemann sum structure</strong></p><p>Rewrite the sum by factoring n from the denominator:</p><p>∑(r/√(n²+r²)) = ∑(r/(n√(1+(r/n)²)))</p><p><strong>Step 2: Express as (1/n)∑f(r/n)</strong></p><p>lim(n→∞) (1/n)∑ᵣ₌₁ⁿ (r/n)/√(1+(r/n)²)</p><p>This is a Riemann sum for f(x) = x/√(1+x²) on [0,1] with partition width Δx = 1/n.</p><p><strong>Step 3: Convert to definite integral</strong></p><p>lim(n→∞) (1/n)∑ f(r/n) = ∫₀¹ x/√(1+x²) dx</p><p><strong>Step 4: Evaluate the integral</strong></p><p>Let u = 1+x², then du = 2x dx</p><p>∫ x/√(1+x²) dx = (1/2)∫ u⁻¹/² du = √u = √(1+x²)</p><p>Evaluating from 0 to 1: [√(1+x²)]₀¹ = √2 - 1</p><p>∴ Answer: <strong>√2 - 1</strong> (Option C)</p>
Correct Answer: C