Definite Integration
Periodic functions and definite integrals
Grade 12
Question:
<p>If \(g(x) = \int_0^x \cos 4t\, dt\), then \(g(x+\pi)\) equals</p>
<p>\(\dfrac{g(x)}{g(\pi)}\)</p>
<p>\(g(x) + g(\pi)\)</p>
<p>\(g(x) - g(\pi)\)</p>
<p>\(g(x) \cdot g(\pi)\)</p>
Step-by-Step Solution
Key Concept: Use the property that g(x+π) = ∫₀^(x+π) cos 4t dt can be split into ∫₀^x cos 4t dt + ∫ₓ^(x+π) cos 4t dt, and recognize that the second integral relates to the periodicity of cosine with period π/2.
<p><strong>Step 1:</strong> Express g(x+π) using the definition:<br/>g(x+π) = ∫₀^(x+π) cos 4t dt</p><p><strong>Step 2:</strong> Split the integral at x:<br/>g(x+π) = ∫₀^x cos 4t dt + ∫ₓ^(x+π) cos 4t dt = g(x) + ∫ₓ^(x+π) cos 4t dt</p><p><strong>Step 3:</strong> Evaluate ∫ₓ^(x+π) cos 4t dt using substitution u = t - x:<br/>∫ₓ^(x+π) cos 4t dt = ∫₀^π cos 4(u+x) du = ∫₀^π cos(4u + 4x) du</p><p><strong>Step 4:</strong> Evaluate [sin(4u + 4x)/4]₀^π:<br/>= [sin(4π + 4x) - sin(4x)]/4 = [sin(4x) - sin(4x)]/4 = 0<br/>(since sin(θ + 4π) = sin θ)</p><p><strong>Step 5:</strong> Therefore:<br/>g(x+π) = g(x) + 0 = g(x)</p><p>∴ Answer: B (which should be g(x))</p>
Correct Answer: B