Explanation:
Start with the series representation for eˣ.
eˣ = ∑ₙ₌₀°° xⁿ / n!
Substitute -x².
e^(-x²) = ∑ₙ₌₀°° (-x²)ⁿ / n!
e^(-x²) = ∑ₙ₌₀°° (-1)ⁿ (x²ⁿ) / n!
Multiply by x².
x² e^(-x²) = x² ∑ₙ₌₀°° (-1)ⁿ (x²ⁿ) / n!
x² e^(-x²) = ∑ₙ₌₀°° (-1)ⁿ (x²ⁿ⁺²) / n!
Integrate both sides (use power rule).
∫ x² e^(-x²) = ∑ₙ₌₀°° (-1)ⁿ (x²ⁿ⁺³) / ((2n + 3) n!)
Evaluate between x=0 and x=0.5.
∫ x² e^(-x²) = ∑ₙ₌₀°° (-1)ⁿ (0.5²ⁿ⁺³) / ((2n + 3) n!)
This is an alternating series, so use alternating series estimation.
(0.5²⁽ⁿ⁺¹⁾⁺³) / ((2(n+1) + 3) (n+1)!) ≤ 0.001
(0.5²ⁿ⁺⁵) / ((2n + 5) (n+1)!) ≤ 0.001
n ≥ 1
So the estimate of the integral is the sum of the first two terms (n=0 and n=1).
I = (-1)⁰ (0.5³) / ((3) 0!) + (-1)¹ (0.5²⁺³) / ((2 + 3) 1!)
I = (0.5³) / 3 − (0.5⁵) / 5
I = 1/24 − 1/160
I = 17/480
I = 0.0354