<p><strong>890.</strong> Let \(L = \lim_{n \to \infty} \dfrac{1}{n^3} \sum_{k=1}^{n} k^2 e^{\frac{k}{n}}\), then find the value of \(e - L\).</p>
Step-by-Step Solution
Key Concept: Convert the sum into a Riemann sum by factoring out n³ and recognizing that ∑k²e^(k/n) corresponds to ∫x²e^x dx over [0,1]. Use the substitution x = k/n with Δx = 1/n.
<p><strong>Step 1: Rewrite as Riemann Sum</strong></p><p>L = lim(n→∞) (1/n³)∑(k=1 to n) k²e^(k/n)</p><p>= lim(n→∞) ∑(k=1 to n) (k/n)²·e^(k/n)·(1/n)</p><p>This is a Riemann sum with x = k/n, Δx = 1/n, and f(x) = x²eˣ</p><p><strong>Step 2: Convert to Definite Integral</strong></p><p>L = ∫₀¹ x²eˣ dx</p><p><strong>Step 3: Integration by Parts (twice)</strong></p><p>For ∫x²eˣ dx:</p><p>Let u = x², dv = eˣ dx</p><p>du = 2x dx, v = eˣ</p><p>∫x²eˣ dx = x²eˣ - 2∫xeˣ dx</p><p>For ∫xeˣ dx:</p><p>Let u = x, dv = eˣ dx</p><p>du = dx, v = eˣ</p><p>∫xeˣ dx = xeˣ - ∫eˣ dx = xeˣ - eˣ</p><p><strong>Step 4: Combine Results</strong></p><p>∫x²eˣ dx = x²eˣ - 2(xeˣ - eˣ) = x²eˣ - 2xeˣ + 2eˣ = eˣ(x² - 2x + 2)</p><p><strong>Step 5: Evaluate from 0 to 1</strong></p><p>L = [eˣ(x² - 2x + 2)]₀¹</p><p>= e¹(1 - 2 + 2) - e⁰(0 - 0 + 2)</p><p>= e(1) - 1(2)</p><p>= e - 2</p><p><strong>Step 6: Find e - L</strong></p><p>e - L = e - (e - 2) = 2</p><p>∴ Answer: <strong>2</strong></p>
Correct Answer: 2