Why Sample Locations Alone Do Not Determine an Estimate
A rectangle estimate combines two kinds of information. The sample location tells us which rate value to use, while the interval width tells us how long that rate is counted for. If two estimates use the same sample locations but assign different widths to them, they can give different totals.
This matters whenever a rate varies over time or across another input. A high rate applied across a wide interval contributes more to the estimate than that same rate applied across a narrow interval. Comparing only the sampled rates leaves out part of the calculation.
Writing an Estimate with Its Widths
Suppose an interval from \(a\) to \(b\) is divided into consecutive subintervals. Let the width of subinterval \(i\) be \(w_i\), and let \(c_i\) be the sample location used for that subinterval. The rectangle estimate is found by multiplying each sampled rate by its own width, then adding the products.
Each width is the right endpoint minus the left endpoint of its subinterval. The widths together cover the whole interval, so \(w_1+w_2+\cdots+w_n=b-a\). If \(f\) is a rate measured in quantity per unit of input, multiplying \(f(c_i)\) by \(w_i\) gives an estimated quantity. The units of the width therefore matter as much as the sampled rate values.
Same Samples, Different Widths
When two estimates use the same sample locations, the rate values at those locations are the same. However, each rate value may be multiplied by a different width. To compare the estimates, list the sample points in order and keep track of the width attached to each one.
For two partitions with the same ordered sample locations \(c_1,\ldots,c_n\), write their widths as \(w_1,\ldots,w_n\) and \(v_1,\ldots,v_n\). Their estimates are
Subtracting the second expression from the first gives a direct comparison. The sampled rates stay the same, while the differences in widths determine how the terms change.
This identity is useful because it shows exactly where a difference between the estimates comes from. An interval whose width increases contributes more of its sampled rate; an interval whose width decreases contributes less. The changes can offset one another, so different widths do not guarantee different estimates. Their effects depend on the rates attached to them.
Worked Example: Same Samples, Different Widths
Worked Example: Same Samples, Different Widths
Consider \(f(x)=x^2\) on the interval from \(0\) to \(4\). Both estimates use the sample locations \(0.5\), \(2\), and \(3.5\). In the first partition, the subintervals are \([0,1]\), \([1,3]\), and \([3,4]\), with widths \(1\), \(2\), and \(1\). Each sample lies in its corresponding subinterval.
The sampled rates are
Weight each sampled rate by its interval width:
For the second partition, use \([0,1.5]\), \([1.5,2.5]\), and \([2.5,4]\). The same three samples lie in the corresponding subintervals, but the widths are now \(1.5\), \(1\), and \(1.5\). Thus
The estimates differ by \(22.75-20.5=2.25\), even though the sample locations and sampled rates are unchanged. The first and third sampled rates receive more weight in the second partition; the middle sampled rate receives less. The increased contributions outweigh the decrease. Both sets of widths total \(4\), as required to cover the same interval.
Why a Constant Rate Is a Special Case
If the rate is constant, the location of a sample does not change the value being multiplied by a width. Suppose \(f(x)=k\) at every sample location. Then an estimate over a fixed interval is \(k\) times the sum of its widths. Since the widths cover the whole interval, that sum is \(b-a\). Therefore every such estimate has the same value \(k(b-a)\), even if the widths differ.
The reason is not that the individual widths are equal; they need not be. It is that every width is multiplied by the same rate, so only their total matters.
Worked Example: Different Widths at a Constant Rate
Take a constant rate \(f(x)=7\) over the interval from \(2\) to \(8\). One partition has subintervals \([2,3]\), \([3,5]\), and \([5,8]\), with widths \(1\), \(2\), and \(3\). Use sample locations \(2.5\), \(4.5\), and \(6.5\). Its estimate is
A second partition has subintervals \([2,4]\), \([4,5]\), and \([5,8]\), with widths \(2\), \(1\), and \(3\). The same sample locations still lie in the corresponding intervals, and each sampled rate is still \(7\):
In each case the widths total \(1+2+3=6\), the length of the interval from \(2\) to \(8\). Checking the result using the constant-rate formula gives \(7(8-2)=7(6)=42\), matching both estimates.
Sample Placement Also Matters
The width comparison holds the sample locations fixed so that the effect of changing widths is visible. But in other comparisons, the widths may stay fixed while the samples move. Then the sampled rate values can change, particularly when the rate varies across the interval. A fair comparison records both pieces of information: which rate is sampled and the width to which it is applied.
Worked Example: Same Widths, Different Samples
Consider \(f(x)=x^2\) on \([0,3]\), divided into three intervals of width \(1\): \([0,1]\), \([1,2]\), and \([2,3]\). First use sample locations \(0.5\), \(1.5\), and \(2.5\). The sampled rates are \(0.25\), \(2.25\), and \(6.25\), so
Now use sample locations \(0.25\), \(1.25\), and \(2.25\), one inside each of the same intervals. The sampled rates are \(0.0625\), \(1.5625\), and \(5.0625\), giving
The widths are unchanged, but the estimates differ because the sampled rates changed. Adding the terms verifies the totals: \(0.25+2.25+6.25=8.75\), while \(0.0625+1.5625+5.0625=6.6875\). This comparison isolates the effect of sample placement rather than the effect of changing widths.
A Complete Comparison Method
When comparing two accumulation estimates, it helps to arrange the information by corresponding interval. The procedure below works directly from the definition of a rectangle estimate: multiply each sampled rate by its interval width and add.
Record the left and right endpoints, or at least the width, for each rectangle in each estimate.
For each subinterval, identify its sample location and the corresponding rate value.
Multiply each sample value by the width of its own subinterval. Do not attach a width to a neighboring sample by mistake.
Calculate each total separately, then subtract or compare the totals. Check that the widths in each partition add to the full interval length.
This method also helps explain a result. If the sample locations are the same, differences arise from the widths attached to the sampled rates. If the widths are the same, differences arise from the sampled rates. If both change, both effects must be included; looking at only one of them cannot determine the comparison in general.
Common Mistake: Adding Rates Without Weighting Them
A tempting but incorrect shortcut is to add the sampled rates and treat that total as the rectangle estimate. In the first worked example, this would give
One might then claim that both partitions have the same estimate because they use the same samples. But \(16.5\) counts each sampled rate as though its width were \(1\). The first partition actually has widths \(1\), \(2\), and \(1\), and the second has widths \(1.5\), \(1\), and \(1.5\). The correct estimates are \(20.5\) and \(22.75\), respectively.
Why This Comparison Matters
Rectangle estimates are designed to represent accumulation across an interval. The sample value describes a rate at one location; the width says how much of the input interval that rate represents in the estimate. Refining or changing intervals can alter the weights, and moving samples can alter the values. Keeping these roles separate makes comparisons more reliable and prevents an estimate from being judged by its sample values alone.
Different partitions of the same interval need not give the same estimate, even when their sample locations match. Nor does a changed estimate by itself tell us which one is closer to the actual accumulation. The calculation here establishes how the estimates are formed and compared; conclusions about their accuracy require additional information about the rate and how it behaves between sample locations.
Check Your Understanding
Use the relationship between sampled rates and interval widths to answer these questions.
- Two estimates use identical sample locations but assign different widths to them. What information is needed to calculate and compare the estimates?
- Why does a constant rate give the same estimate for different partitions of the same interval?
- For a rate \(f\) and widths \(w_1,w_2\), write the rectangle estimate using sample locations \(c_1,c_2\).
- Can changing widths leave an estimate unchanged? Explain how that can happen.
- If two estimates use equal widths but different sample locations, which part of the calculation may change?
