Definite Integration
Limit as a definite integral
Grade 12
Question:
<p>Evaluate: \( L = \lim_{n \to \infty} \sum_{k=0}^{n-1} \frac{k}{n} \left[ \left(\frac{k+1}{n}\right)^{\frac{1}{m}} - \left(\frac{k}{n}\right)^{\frac{1}{m}} \right] \). Find the value of \(m\) if \(L = \dfrac{1}{10}\).</p>
Step-by-Step Solution
Key Concept: Recognize this limit as a Riemann sum for integration by parts: ∫₀¹ x · d(x^(1/m)) dx. Use integration by parts with u = x and dv = x^(1/m-1)/m dx to convert the telescoping sum into an evaluable integral.
<p><strong>Step 1: Recognize the Riemann sum structure</strong></p><p>The sum is ∑_{k=0}^{n-1} (k/n)[(k+1)/n)^(1/m) - (k/n)^(1/m)]</p><p>Let x_k = k/n, Δx = 1/n. This represents: ∑ x_k [f(x_{k+1}) - f(x_k)] where f(x) = x^(1/m)</p><p><strong>Step 2: Recognize as integration by parts</strong></p><p>This is a Riemann sum for the integral ∫₀¹ x df(x) where f(x) = x^(1/m)</p><p>By integration by parts: ∫₀¹ x df(x) = [xf(x)]₀¹ - ∫₀¹ f(x)dx</p><p><strong>Step 3: Evaluate the integration by parts</strong></p><p>= [x · x^(1/m)]₀¹ - ∫₀¹ x^(1/m)dx</p><p>= 1 - ∫₀¹ x^(1/m)dx</p><p><strong>Step 4: Compute the integral</strong></p><p>∫₀¹ x^(1/m)dx = [x^(1/m + 1)/(1/m + 1)]₀¹ = 1/(1/m + 1) = m/(m+1)</p><p><strong>Step 5: Find L and solve for m</strong></p><p>L = 1 - m/(m+1) = (m+1-m)/(m+1) = 1/(m+1)</p><p>Given L = 1/10:</p><p>1/(m+1) = 1/10</p><p>∴ m + 1 = 10</p><p><strong>∴ m = 9</strong></p>
Correct Answer: 9