Refining Partitions and Approximation Accuracy

Why Partition Width Matters

A Riemann sum estimates accumulated change by adding rectangle areas. Each rectangle’s area is its width multiplied by a function value chosen from that subinterval. If the function changes substantially across a wide subinterval, one chosen height may represent the function poorly. Dividing the interval into narrower pieces gives each rectangle a smaller region to represent.

That is the basic reason finer partitions can support more accurate approximations. The careful wording is important, though: a finer partition improves the accuracy we can guarantee for a continuous function when the widths are sufficiently small, but a particular refinement and choice of sample points need not produce a sum closer to the exact value than a previous sum.

Partitions, Refinements, and Mesh

A refinement breaks some or all of the original subintervals into smaller pieces. Its mesh cannot be larger than the original mesh. A partition with a smaller mesh is finer in the sense that no individual subinterval is as wide, even if it is not a refinement that retains all the original points.

For a partition \(P\), choose one sample point \(c_i\) from each subinterval. The corresponding Riemann sum is

The exact accumulated value is the definite integral \(I=\int_a^b f(x)\,dx\). The absolute error of the chosen sum is \(|I-S(P)|\). A more informative question than whether one sum happens to be close is: how small can we guarantee the error will be for any allowed sample points when the mesh is small?

A Coarse Sum and Its Refinements

This example shows one common pattern: for a monotone function, familiar endpoint sums can approach the exact value as the subinterval widths decrease. But the example does not establish that every refinement, using every possible choice of sample points, must reduce the error. In fact, a coarse sum can sometimes be exact by coincidence.

Why Finer Meshes Give a Guarantee

Although an individual refinement can disappoint, continuous functions on a closed interval have a strong accuracy property. On each small subinterval, the function’s values cannot vary much if the subinterval is sufficiently narrow. The chosen sample height, and therefore its rectangle area, must then be close to the function’s accumulated value on that same subinterval.

Here is a justification of the guarantee. On each subinterval \([x_{i-1},x_i]\), continuity ensures that \(f\) attains a minimum \(m_i\) and maximum \(M_i\). The integral over that subinterval lies between \(m_i\Delta x_i\) and \(M_i\Delta x_i\). The sampled rectangle area \(f(c_i)\Delta x_i\) lies between those same bounds. Therefore their difference in absolute value is no greater than \((M_i-m_i)\Delta x_i\).

Adding the subinterval contributions and applying the triangle inequality gives a bound for the total error:

Because \(f\) is continuous on the closed interval \([a,b]\), it is uniformly continuous there. This means that for every \(\eta>0\), some \(\delta>0\) ensures \(|f(x)-f(y)|<\eta\) whenever \(x,y\in[a,b]\) satisfy \(|x-y|<\delta\). Choose \(\eta=\frac{\varepsilon}{b-a}\), assuming \(a<b\). If the mesh is less than the resulting \(\delta\), then any two points in one subinterval are less than \(\delta\) apart, so \(M_i-m_i\leq\eta\). Consequently,

Using a slightly smaller choice of \(\eta\), if needed, makes the final bound strictly less than \(\varepsilon\). The essential point is that the bound holds for every sample point in every subinterval once the mesh is sufficiently small.

Comparing Successive Right Sums

What Refinement Does—and Does Not—Tell You

A smaller mesh limits how much of the function’s variation can be hidden inside any single subinterval. For continuous functions, that produces the accuracy guarantee proved above. It does not, by itself, specify whether a sum is an overestimate or an underestimate, nor does it guarantee that each successive sum’s error is smaller. Such conclusions depend on the function, the sample-point rule, and sometimes additional information about the graph.

For instance, the midpoint sum for \(f(x)=x\) on \([0,1]\) was exact with one interval, while the refined left sum was not. Also, “finer” is not a complete description of an estimation method: to compare sums fairly, state the partition and the sample point used in each subinterval. Changing both at once can change the result in ways that are not captured by width alone.

A Common Mistake: Confusing Width and Height

A frequent error is to assign a rectangle the width of a neighboring subinterval instead of its actual width. In the one-interval midpoint example for \(f(x)=x\) on \([0,1]\), the interval is the whole interval \([0,1]\), so its width is \(1\), not \(\frac12\). The midpoint height is \(\frac12\), giving \(1\cdot\frac12=\frac12\). Writing \(\frac12\cdot\frac12=\frac14\) incorrectly uses half the interval’s width and changes an exact estimate into a false one.

A second version of the same mistake is to claim that refinement must improve every estimate. In the example, the exact coarse midpoint sum is \(\frac12\), whereas the refined left sum is \(\frac14\), with error \(\frac14\). The correction is to distinguish a general guarantee for sufficiently small mesh from a claim about a particular pair of sums. Always compute the actual widths, identify each sample point, and compare both sums with the exact value when it is known.