<p>Let \(f: R \to R\) and \(g: R \to R\) be continuous functions. Then the value of \(\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] dx\) is</p>
Step-by-Step Solution
Key Concept: The integrand is a product of an even function [f(x) + f(-x)] and an odd function [g(x) - g(-x)]. The product of an even and odd function is odd, so its integral over a symmetric interval [-a, a] is always zero.
<p><strong>Step 1:</strong> Identify the parity of each factor.</p><p>Let h(x) = [f(x) + f(-x)] and k(x) = [g(x) - g(-x)]</p><p><strong>Step 2:</strong> Verify h(x) is even.</p><p>h(-x) = [f(-x) + f(x)] = [f(x) + f(-x)] = h(x) ✓ Even function</p><p><strong>Step 3:</strong> Verify k(x) is odd.</p><p>k(-x) = [g(-x) - g(x)] = -[g(x) - g(-x)] = -k(x) ✓ Odd function</p><p><strong>Step 4:</strong> Apply the property that even × odd = odd.</p><p>The product h(x)·k(x) is odd because h(-x)·k(-x) = h(x)·(-k(x)) = -h(x)·k(x)</p><p><strong>Step 5:</strong> Integrate odd function over symmetric interval.</p><p>For any odd function F(x): ∫₍₋ₐ₎ᵃ F(x)dx = 0</p><p>∴ ∫₍₋π/₂₎^(π/2) [f(x) + f(-x)][g(x) - g(-x)]dx = <strong>0</strong></p>
Correct Answer: D