Audit numerical integration loops and error enclosures

Use the existing area setup first

The companion area lesson identifies intersections, sign changes and strip directions. A correct quadrature routine cannot repair a wrongly specified region. This audit uses positive f(x)=3+2x² on [0,2], so the target is both a signed integral and geometric area

Count strips rather than table nodes

With n equal strips and h=(b−a)/n, use left index j=0,...,n−1. Strip j runs from a+jh to a+(j+1)h. There are n+1 nodes but n strips. An inclusive upper index n adds a further strip outside the target interval

Trace an extra endpoint iteration

For the stated example with n=2 and h=1, the correct trapezia contribute 4 and 8, giving 12. A loop j=0,...,2 inclusive adds the strip [2,3], contributing 16, and prints 28. The extra 16 is outside-domain accumulation; it is not the ordinary quadrature error on [0,2]

How many strips does j=0,...,2 inclusive accumulate?

For f(x)=3+2x², an inclusive loop adds trapezia over [0,1], [1,2] and [2,3], contributing 4, 8 and 16. What does it print?

Which correction makes the loop cover the target interval?

Derive an enclosure from monotonicity

For a continuous increasing function, each left rectangle lies below the curve and each right rectangle above it. Therefore L≤I≤R. This applies to the stated signed integral; our positive example is also an area. These inequalities supply evidence that agreement of two arbitrary estimates does not

Distinguish bound and actual error

Here L=8 and R=16, so T=(L+R)/2=12. The enclosure guarantees |T−I|≤4. The exact integral is [3x+2x³/3] from 0 to 2=34/3, so the actual error is 2/3. The bound 4 is valid but not tight. Convexity separately explains why T lies above I.

For L=8, R=16 and T=(L+R)/2, give the certified bound on |T−I|

For f(x)=3+2x² on [0,2], the trapezium estimate is T=12. Integrate f and give the actual absolute error |T−I|

Run an exact audit using technology

from fractions import Fraction as F import json def audit(a, c, b, n): if not all(type(v) is int for v in (a,c,b,n)) or min(a,c,b,n)<=0: raise ValueError('positive integer inputs required') h=F(b,n) def f(x): return c+a*x*x rows=[] for j in range(n): left=j*h right=(j+1)*h rows.append((left,right,h*(f(left)+f(right))/2)) L=sum((h*f(j*h) for j in range(n)),F(0)) R=sum((h*f((j+1)*h) for j in range(n)),F(0)) T=sum((r[2] for r in rows),F(0)) I=c*b+F(a*b**3,3) return {'n':n,'L':str(L),'R':str(R),'T':str(T),'I':str(I), 'bound':str((R-L)/2),'actual':str(abs(T-I)), 'last_right':str(rows[-1][1]), 'rows':[[str(x) for x in row] for row in rows]} if __name__=='__main__': print(json.dumps([audit(2,3,2,n) for n in (2,4,8)],sort_keys=True))

Read the checked outputs

For n=2,4,8 the trapezium estimates are 12,23/2,91/8. The monotonicity bounds are 4,2,1 and actual errors 2/3,1/6,1/24. All runs use the same target [0,2] and exact integral 34/3. The displayed program uses rational arithmetic; executing this model is not evidence that a learner independently operated a tool

For f(x)=3+2x² on [0,2], use four equal trapezia. Give the exact estimate T.

For n=8, the left and right sums are L=83/8 and R=99/8, with T=91/8. Which claim is supported by this enclosure?

Prove the enclosure and explain refinement, ungraded

Derive L≤I≤R for an increasing continuous function, then prove the midpoint estimate has absolute error at most (R−L)/2. Explain why a finite table alone is insufficient to assert those shape conditions. For f(x)=3+2x² on [0,2], the bounds for n=2,4,8 are 4,2,1 and the actual errors are 2/3,1/6,1/24. Compare how each changes as n doubles

Model: use hypotheses rather than agreement

On each subinterval, monotonicity bounds f by its endpoint heights; integration and addition give L≤I≤R. Any value in this enclosure is at most half its width from its midpoint. A table does not establish monotonicity between nodes. For this quadratic, R−L=ab³/n while T−I=ab³/(6n²): different formulas explain the two refinement rates. These rates are not a claim about every integrand

Audit technology and region choice, ungraded

Run the displayed program, record actual inputs and output, and modify one loop endpoint to reproduce the extra-strip error. Explain why the correct output remains an approximation to a curved-function area. Contrast this with exact integration of the piecewise-linear interpolant. Keep this practical report for human review

Model: distinguish target, trace and approximation

The correct rows partition [0,b] exactly. Changing range(n) to range(n+1) adds one strip ending at b+h. Restore the partition before comparing with the target integral. The trapezium sum exactly integrates its straight-line interpolant but generally approximates the curved function. A supplied model output is not a record of the learner’s own execution

Assessment boundary

Practice grades bounded loop traces and exact enclosure calculations. Region sketches, general proofs, independent technology use and claims for arbitrary functions remain ungraded