Definite Integration
Properties of definite integrals
Grade None
Question:
<p>Given \(f\) and \(g\) are two continuous functions such that \(f(x) = f(a - x)\) and \(g(x) + g(a - x) = 4\). Now, \(I = \int_0^a f(x)g(x)\,dx\). Then \(I\) equals:</p>
<p>\(\int_0^a f(x)\,dx\)</p>
<p>\(4\int_0^a f(x)\,dx\)</p>
<p>\(2\int_0^a f(x)\,dx\)</p>
<p>\(3\int_0^a f(x)\,dx\)</p>
Step-by-Step Solution
Key Concept: Use the symmetry property f(x) = f(a-x) to rewrite the integral by substitution u = a-x, then leverage the constraint g(x) + g(a-x) = 4 to create a solvable system of equations for I.
<p><strong>Step 1:</strong> Start with 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>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), we have f(a-u) = f(u). Therefore:</p><p>I = ∫₀ᵃ f(u)g(a-u)du</p><p><strong>Step 3:</strong> We now have two expressions for I:</p><p>• I = ∫₀ᵃ f(x)g(x)dx</p><p>• I = ∫₀ᵃ f(x)g(a-x)dx</p><p><strong>Step 4:</strong> Add these two equations:</p><p>2I = ∫₀ᵃ f(x)[g(x) + g(a-x)]dx</p><p><strong>Step 5:</strong> Apply the given condition g(x) + g(a-x) = 4:</p><p>2I = ∫₀ᵃ f(x)·4 dx = 4∫₀ᵃ f(x)dx</p><p>∴ <strong>I = 2∫₀ᵃ f(x)dx</strong></p>
Correct Answer: C