Integrate local approximations with an error model

Try

Approximate the area under x² on [0,1]. A single midpoint rectangle gives 1/4, a single trapezoid gives 1/2, while the exact integral is 1/3. One Simpson panel combines endpoint and midpoint values as (1/6)[0+4(1/4)+1]=1/3. Exactness here follows from the polynomial shape, not from a universal superiority of three sampled values.

Predict

Predict how halving a uniform mesh changes midpoint and trapezoid errors for a sufficiently smooth fixed function. Their leading global errors scale like h², so the expected reduction is about four. Simpson’s leading error scales like h⁴, suggesting sixteen. These ratios are asymptotic descriptions under smoothness; they can fail before the leading term dominates or when rounding and singular behavior interfere.

Experiment

Compare the quadratic calculation on two meshes.

Rule	One subinterval or panel	Two equal subintervals
Midpoint	1/4	5/16
Trapezoid	1/2	3/8
Simpson	1/3 using one two-subinterval panel	1/3

Compute each error against 1/3. The Simpson column labels deliberately distinguish a panel from an elementary subinterval; confusing these changes error constants by powers of two.

Understand

For composite midpoint with equal width h, expand f about each midpoint. The integral of the linear term vanishes, while |f″|≤M₂ bounds the remaining local integral by M₂h³/24. Summing gives (b−a)h²M₂/24. For the trapezoid rule, integrate the linear interpolant through endpoints. Its remainder has magnitude at most M₂|(x−aᵢ)(x−bᵢ)|/2, whose integral is M₂h³/12. The composite bound is therefore (b−a)h²M₂/12. Both statements require f∈C² on the interval and concern truncation error before floating-point evaluation.

Apply

For even n, composite Simpson uses h=(b−a)/n and weights h/3 times endpoints, alternating interior weights four and two. Each panel spans 2h. It integrates cubics exactly; under f∈C⁴ its absolute error is bounded by (b−a)h⁴M₄/180. One way to justify the coefficient is the local Peano-kernel remainder: the panel functional annihilates degrees zero through three, its fourth-derivative kernel has one sign and absolute integral (2h)⁵/2880. Summing n/2 panels gives the stated constant. A simple quadratic interpolation remainder alone does not transparently establish this fourth-order cancellation; the extra exactness must be accounted for.

The Simpson kernel can be checked explicitly on [−1,1]. For 0≤t≤1, applying the error functional to (x−t)₊³/6 gives K(t)=−(1−t)³(1+3t)/72, and symmetry gives K(−t)=K(t). Its sign is nonpositive and its absolute integral is 1/90. Scaling an interval of length ℓ from [−1,1] multiplies the fourth-derivative error by (ℓ/2)⁵, giving ℓ⁵/2880. The integral representation follows by integrating the fourth derivative four times and using exactness on the cubic Taylor polynomial. This supplies the missing fourth-order mechanism rather than relying only on a coefficient quoted from memory.

Continue

Design tests that integrate constants and low-degree monomials exactly where promised, then test a known smooth nonpolynomial integral across mesh sizes. Report function evaluations, not only panels, since neighboring panels reuse endpoints. A rigorous bound needs a derivative bound that may be unavailable. The next lesson uses refinement differences as practical estimators while making clear that an estimator is not automatically a certificate, especially for narrow peaks, discontinuities or noisy function values.

Review calculus and linear algebra when a prerequisite needs practice. Continue to Refine where evidence says the integral is difficult.