Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>If \(g(x) = \int_0^x \cos^4 t\,dt\) then \(g(x+\pi)\) equals</p>
<p>(a) \(g(x) + g(\pi)\)</p>
<p>(b) \(g(x) - g(\pi)\)</p>
<p>(c) \(g(x) \cdot g(\pi)\)</p>
<p>(d) \(\dfrac{g(x)}{g(\pi)}\)</p>
Step-by-Step Solution
Key Concept: Use the periodicity of cos⁴t and the property that integrating a periodic function over one complete period gives a constant. Split the integral g(x+π) = ∫₀^(x+π) cos⁴t dt into ∫₀^π cos⁴t dt + ∫_π^(x+π) cos⁴t dt, then substitute u = t-π in the second integral.
<p><strong>Step 1:</strong> Write g(x+π) using the definition:</p><p>g(x+π) = ∫₀^(x+π) cos⁴t dt</p><p><strong>Step 2:</strong> Split the integral at t = π:</p><p>g(x+π) = ∫₀^π cos⁴t dt + ∫_π^(x+π) cos⁴t dt</p><p><strong>Step 3:</strong> In the second integral, substitute u = t - π, so dt = du. When t = π, u = 0; when t = x+π, u = x:</p><p>∫_π^(x+π) cos⁴t dt = ∫₀^x cos⁴(u+π) du</p><p><strong>Step 4:</strong> Use the identity cos(u+π) = -cos(u), so cos⁴(u+π) = cos⁴(u):</p><p>∫_π^(x+π) cos⁴t dt = ∫₀^x cos⁴u du = g(x)</p><p><strong>Step 5:</strong> Calculate ∫₀^π cos⁴t dt using cos⁴t = (3 + 4cos(2t) + cos(4t))/8:</p><p>∫₀^π cos⁴t dt = ∫₀^π (3 + 4cos(2t) + cos(4t))/8 dt = [3t/8 + sin(2t)/4 + sin(4t)/32]₀^π = 3π/8</p><p><strong>∴ Answer: g(x+π) = g(x) + 3π/8</strong> (Option A)</p>
Correct Answer: A