Estimate an integral and justify an error bound

Try

An integrable function may have no convenient elementary primitive, or it may be supplied through measurements. Numerical quadrature estimates its integral by combining sampled values. Before computing, distinguish the exact integral of a specified mathematical function from the uncertain physical quantity a fitted function represents.

Predict

For x² on [0,2], predict whether trapezoids overestimate or underestimate. The convex curve lies below each endpoint chord, so trapezoids overestimate. Predict what midpoint rectangles do, then test both rules against the known exact integral 8/3. A known reference makes the error inspectable.

Experiment

Live experiment: from slices to an integral

Predict whether left rectangles underestimate x² on this interval. Compare two slices with four, then switch the rule. Record the signed error, not just whether two rounded estimates look alike.

Function x² on [0, 2]x³ − x on [−1, 1] Rule Left rectanglesMidpoint rectanglesTrapezoids Number of equal slices 24681224 Reset experiment

Width: Δx = (2 − (0)) / 2 = 1.

Estimate: 1. Exact signed integral: 8/3. Estimate minus exact: -1.666667.

Inspect every contribution

Interval	Sample height	Signed contribution
[0, 1]	0	0
[1, 2]	1	1

Displayed decimal estimates and contributions are rounded to six places. These are known smooth teaching functions. A finite estimate is not a general proof of convergence. The blue curve uses 160 drawing segments; panel contributions use the model values. Settings are temporary and do not add completion credit.

Choose a curve and rule in the temporary investigation, record the displayed interval and slice count, then change only the slice count. Explain the sign of the displayed error from the curve and shapes. The table below is a separate worked example on [0,2], not a claim that the live controls use this same interval.

Rule with two equal subintervals	Formula from values 0,1,4	Result
Trapezoid	(1/2)(0+2·1+4)	3
Midpoint	(1/2)²+(3/2)²	5/2
Simpson	(1/3)(0+4·1+4)	8/3

Simpson is exact here because it integrates quadratic interpolants, and the function is itself quadratic. This example does not imply exactness for arbitrary functions. The separate quadrature lab provides another bounded comparison with saved experimental evidence.

Understand

For n equal subintervals of width h=(b−a)/n, the composite trapezoid rule assigns half weight to endpoint values and full weight to interior values, then multiplies by h. The midpoint rule evaluates each subinterval midpoint and multiplies the sum by h. Composite Simpson uses an even n: weights 1 at the endpoints, alternating 4 and 2 inside, and an overall factor h/3. Each Simpson panel spans two subintervals; using the standard formula with odd n is invalid.

Under sufficient smoothness, derivative bounds control worst-case error. If the second derivative is continuous and its magnitude is at most M₂ throughout [a,b], trapezoid error magnitude is at most M₂(b−a)³/(12n²), while midpoint error is at most M₂(b−a)³/(24n²). With a continuous fourth derivative bounded by M₄, composite Simpson error is at most M₄(b−a)⁵/(180n⁴). These formulas assume the stated equal partition and, for Simpson, even n.

For x² on [0,2], M₂=2 and n=2 give trapezoid bound 1/3. The actual error is 3−8/3=1/3, so the bound is attained. A bound is not generally the actual error. A small observed change after refinement is useful evidence but is not automatically a guaranteed error estimate, especially if unresolved oscillation is missed by both grids.

Smoothness assumptions must be checked. The function |x| has a corner at zero and lacks the fourth derivative required by the Simpson bound across that point. Splitting at zero yields two linear pieces that Simpson integrates exactly. A failed error theorem does not make integration impossible; it calls for a different partition or justification.

Quadrature error, floating-point roundoff and model error are distinct. Refining a grid can reduce discretization error while subtraction, evaluation error or missing observations remain. An exact integral in this investigation is an analytic value for its deliberately chosen teaching function, not an exact measurement of a real system.

Apply

Compute the practice sums explicitly, then state the hypotheses behind each claimed bound. In the nonsmooth example, propose a valid split. In the measured-rate question, explain why a more precise numerical result cannot recover behavior that was never observed.

Continue

This integral-calculus unit provides substantive methods and applications, not complete AP or Calculus II preparation. Further series, parametric and polar calculus, arc length, surface area, differential equations and cumulative review remain distinct requirements. Preserve the difference between completing guided work and demonstrating unfamiliar transfer.