Nonuniform Partitions and Unequal Widths

Why Unequal Widths Matter

A partition does not have to divide an interval into equally spaced pieces. One subinterval may be short, another longer, and a third longer still. When we use rectangles to represent accumulation over such a partition, their widths must reflect the actual pieces of the interval.

The height of each rectangle comes from a function value chosen within its subinterval. Its signed contribution is therefore the sampled function value multiplied by that subinterval’s width. Using a different width can change the contribution, even when the sample value itself is unchanged.

Partitions and Their Widths

Let \(a<b\). A partition of \([a,b]\) is a finite, strictly increasing list of points that starts at \(a\) and ends at \(b\):

The points \(x_0,\ldots,x_n\) divide the interval into \(n\) subintervals, \([x_{i-1},x_i]\), for \(i=1,\ldots,n\). The width of the \(i\)th subinterval is defined by subtracting its left endpoint from its right endpoint:

Each width is positive. In a uniform partition, all widths are equal. In a nonuniform partition, at least two widths differ. For a nonuniform partition, there is no single common subinterval width to use in every rectangle.

To form a tagged partition, choose one sample point \(\xi_i\) in each subinterval, so that \(x_{i-1}\le \xi_i\le x_i\). The sample point determines the rectangle’s height through the function value \(f(\xi_i)\).

The factor paired with \(f(\xi_i)\) is the width of the same interval from which \(\xi_i\) was chosen. The widths may differ from one term to the next.

Why Each Value Gets Its Own Width

Over the \(i\)th subinterval, the rectangle has horizontal width \(\Delta x_i\) and signed height \(f(\xi_i)\). Its signed area is height times width, \(f(\xi_i)\Delta x_i\). Adding the rectangles gives the total signed rectangle sum.

This pairing is also required by the structure of the partition. Each subinterval represents a different portion of the original interval, and its contribution must account for how much horizontal distance that portion covers. A sample value from a wide interval is multiplied by a wider amount than the same value from a narrow interval.

The widths themselves add to the length of the whole interval. Indeed, writing out the differences gives a telescoping sum:

Every interior endpoint appears once with a positive sign and once with a negative sign, so those terms cancel. This confirms that the subinterval widths together cover exactly the original interval. Weighting each sample by its own width respects this decomposition.

Worked Sums with Unequal Widths

A Common Shortcut That Can Fail

When a partition has \(n\) subintervals, its average width is \((b-a)/n\). It can be tempting to replace every actual width by that average. This can change the rectangle sum and can give a wrong result: the average width is not generally the width of any particular subinterval.

Return to the first example. The widths were \(1,2,1\), and the sampled values were \(1,3,4\). Their average width is \((4-0)/3=\frac43\). The incorrect substitution would give:

But the required sum is \(1(1)+3(2)+4(1)=11\), not \(\frac{32}{3}\). The difference is that the middle value \(3\) belongs to a width of \(2\), while each other value belongs to a width of \(1\). The average-width calculation assigns all three values the same width instead.

There are cases in which the two calculations coincide. For example, if every sampled value is a constant \(c\), then the original sum is \(c\sum_{i=1}^n\Delta x_i=c(b-a)\). Replacing all widths by the average gives \(c\frac{b-a}{n}n=c(b-a)\) as well. This special agreement does not make average widths valid in general; it occurs because all the sampled values are equal.

A Reliable Setup and Check

For a nonuniform tagged partition, a short setup routine helps prevent misplaced widths and arithmetic errors:

Why This Construction Is Useful

Unequal partitions let a calculation use shorter intervals in portions where more detail is wanted and longer intervals elsewhere. Whatever the choice of partition, the sum still has one consistent meaning: each sampled height represents its own interval, so its signed rectangle contribution is height times that interval’s width.

This construction is the basic way to write rectangle estimates when interval widths differ. A correct sum must preserve the pairing between each sample point and its subinterval. Once the partition points, tags, and widths are written clearly, the calculation is a finite sum of explicit products.