How Smaller Subintervals Change Estimates

Why Subinterval Width Matters

A rectangle estimate uses sampled function values to approximate accumulated change over an interval. Each sampled value is multiplied by the width of the subinterval it represents. If the interval is split into smaller pieces, the estimate uses more samples, and each sample is assigned a smaller width.

The effect of this change is not always the same. For an increasing rate, halving every subinterval raises a left-endpoint estimate and lowers a right-endpoint estimate. For nonmonotone data, either estimate may rise or fall. The added samples matter, but their effect depends on how the function changes across the interval.

Coarse and Refined Endpoint Estimates

Suppose the interval is \([a,b]\), divided into \(n\) equal subintervals, each of width \(h=(b-a)/n\). Write the endpoints as \(x_j=a+jh\), for \(j=0,1,\ldots,n\). The left-endpoint estimate uses the function value at the start of each subinterval; the right-endpoint estimate uses the value at its end.

For the refinement considered here, bisect every coarse subinterval. Its new sample is the midpoint \(m_j=(x_{j-1}+x_j)/2\), and each refined subinterval has width \(h/2\). Thus the refined estimates use both the old endpoint and the new midpoint within each coarse subinterval:

The notation \(2n\) indicates that the number of equal subintervals has doubled. The interval itself has not changed. In each pair of refined subintervals, the new sample \(f(m_j)\) appears alongside the original endpoint values, but all the samples now carry width \(h/2\), not \(h\).

What the Added Samples Change

Subtracting the coarse left estimate from the refined left estimate shows exactly how much the left estimate changes. For each coarse subinterval, the old left sample keeps half its former weight, while the new midpoint sample contributes the other half:

The corresponding change in the right estimate is:

These formulas provide a direct test of the effect of refinement. In the left estimate, compare each new midpoint value with the old left-endpoint value. In the right estimate, compare it with the old right-endpoint value. The width factor \(h/2\) is positive, so the signs of these differences determine whether the estimates rise or fall.

How Monotonicity Determines the Direction

A function is increasing on an interval if \(x<y\) implies \(f(x)\leq f(y)\). It is decreasing if \(x<y\) implies \(f(x)\geq f(y)\). On each coarse subinterval, the midpoint lies between the endpoints. Therefore, for an increasing function, \(f(x_{j-1})\leq f(m_j)\leq f(x_j)\). For a decreasing function, the inequalities reverse.

For an increasing function, each term \(f(m_j)-f(x_{j-1})\) in the left-estimate change is nonnegative, so \(L_{2n}-L_n\geq 0\). Each term \(f(m_j)-f(x_j)\) in the right-estimate change is nonpositive, so \(R_{2n}-R_n\leq 0\). For a decreasing function, those term-by-term signs reverse, which proves both reversed inequalities. This also explains why monotonicity matters: it guarantees the sign of every contribution, not merely the sign of their total.

For an increasing function, the left and right estimates on either grid are ordered from below to above. Combining that ordering with the refinement inequalities gives:

The coarse estimates therefore enclose the refined estimates: the coarse left estimate is no greater than the refined left estimate, and the coarse right estimate is no less than the refined right estimate. The bracket formed by the refined estimates is no wider than the coarse bracket. For a decreasing function, the corresponding ordering is \(R_n\leq R_{2n}\leq L_{2n}\leq L_n\). As established in the earlier material on monotone rates, when these endpoint estimates bound the total accumulation, this narrowing gives tighter bounds.

Worked Examples

Common Pitfall: Assuming Refinement Always Improves in One Direction

A common mistake is to assume that halving the widths must always increase a left estimate and decrease a right estimate. That conclusion is valid for increasing functions, but it does not follow from the smaller widths alone. For the second data set in the previous example, the coarse left estimate is \(0\), while the refined left estimate is \(2\); the coarse right estimate is also \(0\), while the refined right estimate is \(2\). Claiming that the refined right estimate must be smaller would give the wrong comparison.

The correction is to check the function’s behavior before predicting a direction. If monotonicity holds, use the theorem. If it does not, compute the estimates or inspect the change formulas. For example, the left-estimate change depends on the sums of \(f(m_j)-f(x_{j-1})\); these differences need not all have the same sign for nonmonotone data.

Why Refinement Is Useful

When a rate is measured in units per unit of time, multiplying a rate sample by a time width gives an estimate in units of accumulated change. For example, if a rate is in litres per minute and a subinterval width is in minutes, each rectangle contributes an estimated number of litres. Refinement keeps the total time interval fixed but uses shorter time widths and additional rate samples.

For monotone rates, the change in the endpoint estimates is predictable, and the refined estimates provide a tighter bracket for the total accumulation. For irregular or nonmonotone rates, refinement still supplies more sample information, but the direction of change must be determined from the actual values. In either case, recording the interval, the widths, the sample locations, and the rate values helps prevent errors when setting up the estimate.