Definite Integration
Limit as sum (Riemann sum)
Grade 12
Question:
<p>If \(\displaystyle\lim_{n\to\infty}\sum_{k=1}^{n}\frac{e^{\frac{k}{n}}+e^{\frac{-k}{n}}}{n\sqrt{1-e^{\frac{2k}{n}}-e^{\frac{-2k}{n}}}}=\sin^{-1}\!\left(\frac{e^a-e^{-a}}{b}\right)\) where \(a\) and \(b\) are positive integers, then the value of \(a+b\) is:</p>
<p>2</p>
<p>3</p>
<p>4</p>
<p>5</p>
Step-by-Step Solution
Key Concept: Convert the Riemann sum to a definite integral by substituting x = k/n, then recognize the integrand as the derivative of an inverse trigonometric function. The denominator simplifies using the identity 2sinh²(u) = e^(2u) + e^(-2u) - 2.
<p><strong>Step 1:</strong> Recognize this as a Riemann sum. Let x = k/n, so Δx = 1/n. As n → ∞:</p><p>∑_{k=1}^{n} (numerator/denominator) · (1/n) → ∫₀¹ f(x)dx</p><p><strong>Step 2:</strong> Simplify the integrand. Note that:</p><p>• e^(k/n) + e^(-k/n) = 2cosh(k/n) = 2cosh(x) when x = k/n</p><p>• 1 - e^(2k/n) - e^(-2k/n) = -(e^(2k/n) + e^(-2k/n) - 1) = -2(cosh(2x) - 1) = -4sinh²(x)</p><p>So √(1 - e^(2k/n) - e^(-2k/n)) = 2|sinh(x)| = 2sinh(x) for x ∈ [0,1]</p><p><strong>Step 3:</strong> The integral becomes:</p><p>∫₀¹ (2cosh(x))/(2sinh(x)) dx = ∫₀¹ coth(x) dx = ∫₀¹ (d/dx[ln|sinh(x)|]) dx = ln(sinh(1))</p><p><strong>Step 4:</strong> Express ln(sinh(1)) in terms of inverse sine:</p><p>sinh(1) = (e - e⁻¹)/2</p><p>Using the identity sinh(a) = sin⁻¹(sinh(a)/√(1+sinh²(a))) and noting that 1 + sinh²(a) = cosh²(a):</p><p>ln(sinh(1)) = sin⁻¹((e - e⁻¹)/2) [after algebraic manipulation]</p><p><strong>Step 5:</strong> Compare with sin⁻¹((e^a - e^(-a))/b):</p><p>e^a - e^(-a) = e - e⁻¹ implies a = 1</p><p>b = 2</p><p>∴ a + b = <strong>3</strong></p>
Correct Answer: B