Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>Let \(f(x) = \dfrac{e^x}{1+e^x}\). If \(I_1 = \int_{f(-a)}^{f(a)} x\,g\{x(1-x)\}\,dx\) and \(I_2 = \int_{f(-a)}^{f(a)} g\{x(1-x)\}\,dx\), then find the value of \(\dfrac{I_2}{I_1}\).</p>
<p>\(1\)</p>
<p>\(-3\)</p>
<p>\(-1\)</p>
<p>\(2\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = e^x/(1+e^x) satisfies f(-a) + f(a) = 1 (complementary property), making the integration limits symmetric about x = 1/2. Use substitution u = 1-x to relate I₁ and I₂ through this symmetry.
<p><strong>Step 1:</strong> Verify the complementary property of f(x).</p><p>f(x) = e^x/(1+e^x) and f(-x) = e^(-x)/(1+e^(-x)) = 1/(1+e^x)</p><p>Therefore: f(x) + f(-x) = 1, so f(-a) + f(a) = 1</p><p><strong>Step 2:</strong> Let α = f(-a) and β = f(a), where α + β = 1. Rewrite:</p><p>I₁ = ∫_α^β x·g{x(1-x)} dx</p><p>I₂ = ∫_α^β g{x(1-x)} dx</p><p><strong>Step 3:</strong> In I₁, substitute u = 1 - x, so x = 1 - u, dx = -du:</p><p>I₁ = ∫_β^α (1-u)·g{(1-u)u} (-du) = ∫_α^β (1-u)·g{u(1-u)} du</p><p><strong>Step 4:</strong> Since g{u(1-u)} = g{(1-u)u}, we have:</p><p>I₁ = ∫_α^β (1-u)·g{u(1-u)} du = ∫_α^β g{u(1-u)} du - ∫_α^β u·g{u(1-u)} du</p><p>I₁ = I₂ - I₁</p><p><strong>Step 5:</strong> Solving: 2I₁ = I₂</p><p>∴ I₂/I₁ = <strong>2</strong></p>
Correct Answer: D