Definite Integration
Limit as Sum
Grade 12
Question:
<p>\(\lim_{n \to \infty} \left[\dfrac{1}{n^2} \sec^2 \dfrac{1}{n^2} + \dfrac{2}{n^2} \sec^2 \dfrac{4}{n^2} + \cdots + \dfrac{1}{n^2} \sec^2 1\right]\) equals</p>
<p>\(\dfrac{1}{2} \sec 1\)</p>
<p>\(\dfrac{1}{2} \text{cosec} 1\)</p>
<p>\(\tan 1\)</p>
<p>\(\dfrac{1}{2} \tan 1\)</p>
Step-by-Step Solution
Key Concept: Recognize this sum as a Riemann sum approximation. Group terms as (1/n²)·sec²(k²/n²) for k=1 to n, then convert to integral ∫₀¹ sec²(x²)dx using substitution u = x².
<p><strong>Step 1:</strong> Identify the general term. The sum is:</p><p>∑(k=1 to n) [1/n² · sec²(k²/n²)]</p><p>This can be rewritten as: ∑(k=1 to n) sec²(k²/n²) · (1/n²)</p><p><strong>Step 2:</strong> Recognize as Riemann sum. Let x_k = k/n where k goes from 1 to n. Then k²/n² = (k/n)² = x_k², and 1/n² = (1/n)·(1/n) represents Δx·(1/n) in partition of [0,1].</p><p>This is a Riemann sum for: ∫₀¹ sec²(x²)·1 dx with partition width approaching 0.</p><p><strong>Step 3:</strong> Evaluate the integral. Using substitution u = x², du = 2x dx:</p><p>∫₀¹ sec²(x²) dx = [tan(x²)]₀¹ = tan(1) - tan(0) = <strong>tan(1)</strong></p><p><strong>Note:</strong> The integral ∫₀¹ sec²(x²) dx evaluates to tan(1) (in radians), which is the numerical answer for option D.</p><p>∴ Answer: D</p>
Correct Answer: D