Definite Integration
Definite integrals using properties and even functions
Grade 12
Question:
<p>Consider \(I = \displaystyle\int_{-a}^{a} \frac{f(x)}{1+e^{2x+1}}\, dx\), where \(f(x)\) is an even function. Evaluate: \[\int_{-4}^{4} \frac{x^2}{(x^2+16)(1+e^{x^3})}\, dx\]</p>
Step-by-Step Solution
Key Concept: Use the property that for even function f(x), ∫_{-a}^{a} f(x)/(1+e^{2x+1}) dx = ∫_0^a f(x)/(1+e) dx by exploiting the symmetry and the exponential term's behavior. Then recognize x²/(x²+16) is even and 1/(1+e^{x³}) has a special cancellation property when paired with its negative counterpart.
<p><strong>Step 1:</strong> Observe that g(x) = x²/(x²+16) is even. Denote I = ∫₋₄⁴ x²/[(x²+16)(1+e^{x³})] dx</p><p><strong>Step 2:</strong> For the exponential term, note the key property: 1/(1+e^{x³}) + 1/(1+e^{-x³}) = 1</p><p><strong>Step 3:</strong> Split: I = ∫₋₄⁰ g(x)/(1+e^{x³}) dx + ∫₀⁴ g(x)/(1+e^{x³}) dx</p><p><strong>Step 4:</strong> In the first integral, substitute u = -x: ∫₋₄⁰ g(x)/(1+e^{x³}) dx = ∫₀⁴ g(u)/(1+e^{-u³}) du (using g even)</p><p><strong>Step 5:</strong> Add the two integrals: I = ∫₀⁴ g(x)[1/(1+e^{x³}) + 1/(1+e^{-x³})] dx = ∫₀⁴ g(x)·1 dx = ∫₀⁴ x²/(x²+16) dx</p><p><strong>Step 6:</strong> Evaluate: ∫₀⁴ x²/(x²+16) dx = ∫₀⁴ [1 - 16/(x²+16)] dx = [x - 4·arctan(x/4)]₀⁴ = 4 - 4·arctan(1) = 4 - 4(π/4) = 4 - π</p><p><strong>Step 7:</strong> Calculate: 4 - π ≈ 4 - 3.14159 ≈ 0.8571</p><p>∴ Answer: <strong>0.8571</strong> or <strong>4 - π</strong></p>
Correct Answer: 0.8571