Definite Integration
Properties of definite integrals
Grade 12
Question:
<p>Evaluate: \(\displaystyle\int_{-4}^{4} \frac{x^2}{(x^2+16)(1+e^x)}\,dx\) (up to four decimal places).</p>
Step-by-Step Solution
Key Concept: Decompose the integrand using properties of even/odd functions and split it as f(x) + g(x) where g(x) is odd. The odd part integrates to zero over symmetric limits, leaving only the even component to evaluate.
<p><strong>Step 1: Recognize symmetry and split the integrand</strong></p><p>Let I = ∫₋₄⁴ [x²/((x²+16)(1+eˣ))] dx</p><p>Use the key identity: 1/(1+eˣ) + 1/(1+e⁻ˣ) = 1</p><p>Rewrite: x²/((x²+16)(1+eˣ)) = x²/((x²+16)(1+eˣ)) · [1/(1+eˣ) + 1/(1+e⁻ˣ)] / [1/(1+eˣ) + 1/(1+e⁻ˣ)]</p><p><strong>Step 2: Decompose into even and odd parts</strong></p><p>I = ∫₋₄⁴ [x²/((x²+16)(1+eˣ))] dx + ∫₋₄⁴ [x²/((x²+16)(1+e⁻ˣ))] dx] / 2</p><p>Let f(x) = x²/((x²+16)(1+eˣ)) and f(-x) = x²/((x²+16)(1+e⁻ˣ))</p><p>Then: 2I = ∫₋₄⁴ [x²/(x²+16)] · [1/(1+eˣ) + 1/(1+e⁻ˣ)] dx = ∫₋₄⁴ [x²/(x²+16)] dx</p><p><strong>Step 3: Evaluate the simplified integral</strong></p><p>∫₋₄⁴ [x²/(x²+16)] dx = ∫₋₄⁴ [1 - 16/(x²+16)] dx</p><p>= [x - 16·(1/4)arctan(x/4)]₋₄⁴</p><p>= [x - 4arctan(x/4)]₋₄⁴</p><p>= [4 - 4arctan(1)] - [-4 - 4arctan(-1)]</p><p>= [4 - 4(π/4)] - [-4 + 4(π/4)]</p><p>= [4 - π] - [-4 + π] = 8 - 2π</p><p><strong>Step 4: Calculate final answer</strong></p><p>2I = 8 - 2π</p><p>I = 4 - π ≈ 4 - 3.1416</p><p>∴ Answer: <strong>0.8584</strong></p>
Correct Answer: 0