1 · Estimating integrals using strips
Divide [a,b] into n subintervals of width h = (b−a)/n. Approximate the integral by rectangles — left, right, or midpoint. Errors shrink as 1/n (first-order methods), more slowly than Taylor estimates but always available.
2 · The trapezoidal rule
Replace each rectangle by a trapezoid and average left and right ends:
Tₙ = h/2 · [f(x₀) + 2Σ_{i=1}^{n−1} f(xᵢ) + f(xₙ)].
Second-order accurate: error O(1/n²).
3 · Simpson’s rule — parabola-fitting
Fit a parabola through triples of points (n must be even):
Sₙ = (h/3) · [f(x₀) + 4Σ odd i + 2Σ even i (excluding ends) + f(xₙ)]. Works beautifully on cubics (exact!) and usually rivals trapezoids with far fewer strips.
Estimate ∫₀¹ e^{−x²}dx with Simpson, n=2 (h=0.5): values are 1, e^{−0.25}≈0.779, e^{−1}≈0.368. Simpson: 0.5/3·[1 + 4·0.779 + 0.368] ≈ 0.7477. The true value is 0.7468 — error 0.001. Trapezoidal with the same two strips gives 0.7533, with more error. Simpson wins.
4 · The error in our approximations
For smooth f, trapezoidal error ≤ −(b−a)h²/12 · f″(ξ), Simpson error ≤ −(b−a)h⁴/180 · f⁽⁴⁾(ξ). The h⁴ in Simpson explains its reputation for accuracy per strip.
Simpson’s rule demands an even number of strips — an odd n silently breaks the pairing. And on non-smooth functions (sharp corners, kinks), both rules lose their high order: refine near the kink, and don’t trust an error bound that assumed 4th-order smoothness you don’t have.
Sₙ = (h/3)[f(x₀) + 4Σodd + 2Σeven + f(xₙ)] error ≈ −(b−a)h⁴/180 · f⁽⁴⁾(ξ)
Exact for: Simpson on polynomials degree ≤ 3
Choose n for desired digits: more strips are cheap, error bounds are not
Common mistakes
- Misidentifying h: it’s (b−a)/n, not 1/n — scaling matters.
- Applying Simpson with odd n — it needs an even number of subintervals.
- Trusting a trapezoid on a sharp peak; refine where f curves, not where it’s flat.
- Quoting an estimate without its error bound — “≈ 0.75” is a claim; “0.75 ± 0.01” is an answer.