How do you do this question?

Step-by-step 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

the answer is 15% decrease