Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>If \(f(x) = \dfrac{e^x}{1 + e^x}\), \(I_1 = \int_{f(-a)}^{f(a)} x g\{x(1-x)\} \, dx\) and \(I_2 = \int_{f(-a)}^{f(a)} g\{x(1-x)\} \, dx\), then the value of \(\dfrac{I_2}{I_1}\) is</p>
<p>\(2\)</p>
<p>\(-3\)</p>
<p>\(-1\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Recognize that f(x) = e^x/(1+e^x) satisfies f(-x) + f(x) = 1, making the integration limits symmetric about x = 1/2. Use this symmetry property combined with substitution u = 1-x to relate I₁ and I₂.
<p><strong>Step 1:</strong> Verify the symmetry property of f(x).</p><p>f(x) = e^x/(1+e^x). Check: f(-x) = e^(-x)/(1+e^(-x)) = 1/(e^x+1)</p><p>Therefore: f(-x) + f(x) = 1/(1+e^x) + e^x/(1+e^x) = 1</p><p>This means if α = f(-a), then f(a) = 1 - α, so the limits are α and (1-α).</p><p><strong>Step 2:</strong> Express I₁ using the symmetric limits.</p><p>I₁ = ∫[α to 1-α] x·g{x(1-x)} dx</p><p><strong>Step 3:</strong> Apply substitution u = 1-x in I₁.</p><p>When x = α, u = 1-α; when x = 1-α, u = α</p><p>I₁ = ∫[1-α to α] (1-u)·g{(1-u)u} (-du) = ∫[α to 1-α] (1-u)·g{u(1-u)} du</p><p>Note: x(1-x) = (1-u)u is the same function, and the integrand becomes symmetric about x = 1/2.</p><p><strong>Step 4:</strong> Add the two expressions for I₁.</p><p>I₁ = ∫[α to 1-α] x·g{x(1-x)} dx</p><p>I₁ = ∫[α to 1-α] (1-x)·g{x(1-x)} dx</p><p>Adding: 2I₁ = ∫[α to 1-α] [x + (1-x)]·g{x(1-x)} dx = ∫[α to 1-α] g{x(1-x)} dx = I₂</p><p><strong>Step 5:</strong> Calculate the ratio.</p><p>∴ I₂/I₁ = 2</p>
Correct Answer: A