Definite Integration
Properties of Definite Integrals
Grade 12
Question:
<p>Let \(f\) be a continuous and even function such that \(\int_0^a f(x)\,dx = 10\). If \(g(x)\) is a continuous positive function such that \(g(x)g(-x) = 1\) and \(\int_0^a g(x)\,dx = 5\), then find the value of \(\int_{-a}^{a} \dfrac{f(x)}{1+g(x)}\,dx\).</p>
Step-by-Step Solution
Key Concept: Use the even property of f(x) and the condition g(x)g(-x)=1 to split the integral into symmetric parts, then substitute u=-x in one part to create a solvable system.
<p><strong>Step 1:</strong> Since f is even: ∫₋ₐᵃ f(x)dx = 2∫₀ᵃ f(x)dx = 2(10) = 20</p><p><strong>Step 2:</strong> Split the integral: ∫₋ₐᵃ f(x)/(1+g(x)) dx = ∫₋ₐ⁰ f(x)/(1+g(x)) dx + ∫₀ᵃ f(x)/(1+g(x)) dx</p><p><strong>Step 3:</strong> In the first integral, substitute u = -x, so du = -dx. Since f is even, f(-u) = f(u), and g(-u) = 1/g(u):</p><p>∫₋ₐ⁰ f(x)/(1+g(x)) dx = ∫₀ᵃ f(u)/(1+1/g(u)) du = ∫₀ᵃ f(u)·g(u)/(1+g(u)) du</p><p><strong>Step 4:</strong> Add both parts:</p><p>∫₋ₐᵃ f(x)/(1+g(x)) dx = ∫₀ᵃ [f(x)/(1+g(x)) + f(x)·g(x)/(1+g(x))] dx</p><p>= ∫₀ᵃ f(x)·[1+g(x)]/(1+g(x)) dx = ∫₀ᵃ f(x) dx = 10</p><p><strong>Step 5:</strong> Note: The given condition ∫₀ᵃ g(x)dx = 5 is extraneous information used to verify consistency but doesn't affect the final answer through the key symmetry argument.</p><p>∴ <strong>Answer: 10</strong></p>
Correct Answer: 10