From Endpoint Samples to an Accumulation Estimate
When a rate varies over time, knowing a few of its values does not usually tell us the exact accumulated change. We can still build an estimate by dividing the time interval into smaller pieces, sampling the rate at a chosen location in each piece, and multiplying that sampled rate by the piece’s duration.
This tutorial focuses on the choice of sampling location at the right endpoint of each subinterval. The essential bookkeeping is to match each sampled rate with the width of the subinterval that ends at that sample. A mismatch between the sample and the width changes the calculation.
Partitions and Right Endpoints
A partition of an interval \([a,b]\) is a finite increasing list of points that includes its endpoints:
The partition creates \(n\) subintervals. The \(i\)-th subinterval runs from \(x_{i-1}\) to \(x_i\), so its width is
Because the partition points are strictly increasing, every width is positive. The right endpoint of the \(i\)-th subinterval is \(x_i\), not \(x_{i-1}\). If a rate is represented by a function \(r\), the right-endpoint sample on that subinterval is therefore \(r(x_i)\).
The index is important. The \(i\)-th width is \(x_i-x_{i-1}\), and the right-endpoint rate that belongs with it is \(r(x_i)\). The same pairing applies whether the partition has equal or unequal widths.
Why the Product Uses Rate Times Width
A rate describes change per unit of the input variable. If \(r\) is measured in liters per minute and a subinterval lasts \(\Delta t_i\) minutes, then a rate value multiplied by that duration has units of liters. More generally, a rate sampled on a subinterval gives a simple estimate of the change over that subinterval when multiplied by its width.
For the right-endpoint construction, we use the sampled value \(r(x_i)\) as a representative rate on the subinterval from \(x_{i-1}\) to \(x_i\). The estimated contribution from that subinterval is \(r(x_i)\Delta x_i\). Adding these contributions gives an estimate over the whole partition.
The subintervals cover \([a,b]\) without gaps and have disjoint interiors. Adjacent closed subintervals share their boundary point: for instance, \([x_0,x_1]\) and \([x_1,x_2]\) both contain \(x_1\). That shared point does not represent a duration or width, so it is not counted twice in the sum of the subinterval contributions.
This explains how to construct the estimate; it does not claim that the estimate is exact. The sampled rate may differ from the rate at other points in the same subinterval. The right-endpoint sum is built from the information specified by the partition and its right-endpoint samples.
A Reliable Construction Process
List \(x_0,x_1,\ldots,x_n\) from the start of the interval to the end.
For subinterval \(i\), calculate \(\Delta x_i=x_i-x_{i-1}\).
Use the rate at \(x_i\), the endpoint on the right of subinterval \(i\).
Form \(r(x_i)\Delta x_i\) for each subinterval and add all \(n\) terms.
Confirm that each term has the units of the accumulated quantity and that negative rates contribute negative terms.
Worked Examples
Worked Example: Estimating a Flow Accumulation
Water flows into a tank over the time interval from \(t=0\) to \(t=6\) minutes. Use the partition \(0,1,4,6\) and the flow rates at the right endpoints shown below to construct a right-endpoint estimate.
| Right endpoint \(t_i\) (min) | Flow rate \(q(t_i)\) (L/min) |
|---|---|
| 1 | 2 |
| 4 | 3.5 |
| 6 | 1.5 |
The subintervals are \([0,1]\), \([1,4]\), and \([4,6]\). Their widths are
Pair each width with the flow rate at the subinterval’s right endpoint:
The estimated accumulation is \(15.5\) liters. Check the products and units: \(2\text{ L/min}\cdot1\text{ min}=2\text{ L}\), \(3.5\text{ L/min}\cdot3\text{ min}=10.5\text{ L}\), and \(1.5\text{ L/min}\cdot2\text{ min}=3\text{ L}\). Their sum is \(2+10.5+3=15.5\text{ L}\), as stated.
Worked Example: Displacement from Signed Velocity Samples
A moving object has velocity samples \(v(2)=6\) meters per second and \(v(5)=-1\) meter per second. Use the partition \(0,2,5\) to estimate its displacement from \(t=0\) to \(t=5\) seconds with a right-endpoint sum.
There are two subintervals. The first has width \(2-0=2\) seconds and right endpoint \(2\); the second has width \(5-2=3\) seconds and right endpoint \(5\). Thus,
The estimate is \(9\) meters of displacement. The negative velocity on the second subinterval produces a negative contribution, as required for signed change. Check the units and arithmetic: \((6\text{ m/s})(2\text{ s})=12\text{ m}\) and \((-1\text{ m/s})(3\text{ s})=-3\text{ m}\), so \(12+(-3)=9\text{ m}\).
Worked Example: Using an Unequal Partition for Production
A machine’s production rate is recorded at times \(t=3,7,8\) hours. Use the interval from \(t=2\) to \(t=8\), partitioned as \(2,3,7,8\). Suppose the recorded rates at those right endpoints are \(4\), \(2\), and \(5\) items per hour, respectively.
First determine the three widths:
The right-endpoint samples are \(4\) items per hour at \(t=3\), \(2\) items per hour at \(t=7\), and \(5\) items per hour at \(t=8\). The estimate is
The estimate is \(17\) items. To check, the three contributions are \(4\) items, \(8\) items, and \(5\) items; \(4+8+5=17\). Each rate in items per hour was multiplied by a width in hours, leaving items.
A Common Indexing Mistake
A frequent error is to use the right endpoint for one subinterval but multiply it by the width of a neighboring subinterval. Consider the partition \(0,2,5\), with rates \(r(0)=1\), \(r(2)=3\), and \(r(5)=7\). The first width is \(2-0=2\), and its right-endpoint sample is \(r(2)=3\). The second width is \(5-2=3\), and its right-endpoint sample is \(r(5)=7\).
The right-endpoint sum is therefore
A wrong calculation would be \(3(3)+7(2)=9+14=23\). Although it uses both right-endpoint rates, it pairs each with the other subinterval’s width. The correction is to keep the endpoint and width together: \(r(2)\) goes with \(2-0\), and \(r(5)\) goes with \(5-2\). Checking each term against its own subinterval catches this error.
What to Check Before Finishing
Before accepting a right-endpoint sum, check the construction rather than only the final arithmetic.
- Does the partition begin at \(a\), end at \(b\), and list all points in increasing order?
- For each subinterval, is the width calculated as right boundary minus left boundary?
- Is the sample taken at the boundary on the right of that same subinterval?
- Do the units of rate multiplied by width produce the units of accumulation?
- If a rate is negative, does its contribution retain the negative sign?
A right-endpoint sum is a finite estimate assembled from one sampled rate and one width per subinterval. Its defining feature is the consistent pairing \(r(x_i)\) with \(x_i-x_{i-1}\). Once that pairing is clear, the construction works for equal or unequal partitions, positive or signed rates, and many kinds of accumulated quantities.
Check Your Understanding
For each question, identify the subintervals and match every right-endpoint sample with its corresponding width before calculating.
- For the partition \(1,3,6\), write the widths of the two subintervals and identify their right endpoints.
- A rate has values \(r(2)=4\) and \(r(5)=1\). For the partition \(0,2,5\), write the right-endpoint sum without simplifying it.
- If a rate is measured in kilograms per minute and the width is measured in minutes, what are the units of their product?
- For a signed rate, explain why a negative sampled value contributes a negative term to the right-endpoint sum.
- In the partition \(0,1,4\), which width must be paired with the sample at \(t=4\)?
