Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>If \( L = \dfrac{\displaystyle\int_0^{n\pi} e^{-x}(\sin^4 ax + \cos^2 ax)\,dx}{\displaystyle\int_0^{\pi} e^{-x}(\sin^4 ax + \cos^2 ax)\,dx} \), where \( a \in \mathbb{R} \) then:</p>
<p>(a) If \( a = 1 \), then \( \lim_{n\to\infty} L < 1 \)</p>
<p>(b) If \( a = 2 \), then \( \lim_{n\to\infty} L > 1 \)</p>
<p>(c) If \( a = 3 \), then \( \lim_{n\to\infty} L < 1 \)</p>
<p>(d) If \( a = 4 \), then \( \lim_{n\to\infty} L > 1 \)</p>
Step-by-Step Solution
Key Concept: Recognize that the integrand f(x) = e^(-x)(sin⁴(ax) + cos²(ax)) has period π in its trigonometric part, and use the property that ∫₀^(nπ) f(x)dx can be decomposed as a sum of n identical integrals over shifted intervals [kπ, (k+1)π], weighted by exponential decay factors e^(-kπ).
<p><strong>Step 1:</strong> Identify periodicity. Since sin⁴(ax) and cos²(ax) have period π/|a|, their sum has period π. Thus f(x) = e^(-x)(sin⁴(ax) + cos²(ax)) is periodic with period π in the trigonometric part.</p><p><strong>Step 2:</strong> Decompose the numerator integral:<br>∫₀^(nπ) e^(-x)f(x)dx = Σ(k=0 to n-1) ∫_(kπ)^((k+1)π) e^(-x)f(x)dx</p><p><strong>Step 3:</strong> Substitute u = x - kπ in the k-th integral:<br>∫_(kπ)^((k+1)π) e^(-x)f(x)dx = e^(-kπ) ∫₀^π e^(-u)f(u)du = e^(-kπ) ∫₀^π e^(-x)f(x)dx</p><p><strong>Step 4:</strong> Sum the series:<br>Numerator = (∫₀^π e^(-x)f(x)dx)·Σ(k=0 to n-1) e^(-kπ) = (∫₀^π e^(-x)f(x)dx)·(1 - e^(-nπ))/(1 - e^(-π))</p><p><strong>Step 5:</strong> Calculate the ratio:<br>L = [(1 - e^(-nπ))/(1 - e^(-π))] = (e^(nπ) - 1)/(e^π - 1)</p><p>∴ Answer: BD</p>
Correct Answer: BD