Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>If \(f\) and \(g\) are continuous functions in \([0, a]\) satisfying \(f(x) = f(a-x)\) and \(g(x) + g(a-x) = 2\), then prove that<br>\(\int_0^a f(x)\,g(x)\,dx = \int_0^a f(x)\,dx\).</p>
Step-by-Step Solution
Key Concept: Use the property of definite integrals by substituting u = a - x to establish a relationship, then add the original integral to itself to isolate the desired form. The condition g(x) + g(a-x) = 2 ensures the g(x) terms combine elegantly.
<p><strong>Step 1:</strong> Let I = ∫₀ᵃ f(x)g(x)dx. Use substitution u = a - x, so du = -dx.</p><p>When x = 0, u = a; when x = a, u = 0.</p><p>Therefore: I = ∫ₐ⁰ f(a-u)g(a-u)(-du) = ∫₀ᵃ f(a-u)g(a-u)du</p><p><strong>Step 2:</strong> Since f(x) = f(a-x) (given), we have f(a-u) = f(u).</p><p>So: I = ∫₀ᵃ f(u)g(a-u)du = ∫₀ᵃ f(x)g(a-x)dx</p><p><strong>Step 3:</strong> Now add the original integral with this result:</p><p>2I = ∫₀ᵃ f(x)g(x)dx + ∫₀ᵃ f(x)g(a-x)dx</p><p>2I = ∫₀ᵃ f(x)[g(x) + g(a-x)]dx</p><p><strong>Step 4:</strong> Apply the given condition g(x) + g(a-x) = 2:</p><p>2I = ∫₀ᵃ f(x)·2·dx = 2∫₀ᵃ f(x)dx</p><p><strong>Step 5:</strong> Divide both sides by 2:</p><p>I = ∫₀ᵃ f(x)dx</p><p>∴ ∫₀ᵃ f(x)g(x)dx = ∫₀ᵃ f(x)dx ✓</p>
Correct Answer: Proof-based