Why Unequal Widths Matter
When a quantity changes over an interval, a rate tells us how quickly that change occurs. To estimate the change over a short part of the interval, the rate must be considered together with the length of that part. If the parts have different lengths, they do not all receive the same width factor.
For example, a rate of \(3\) litres per minute acting for \(2\) minutes corresponds to a different amount of change than the same rate acting for \(5\) minutes. When an interval is divided into pieces of unequal size, each piece has its own width. The width for one piece must be paired with the rate sample associated with that same piece.
Widths from Consecutive Partition Points
A partition of an interval \([a,b]\) is a finite increasing list of points beginning at \(a\) and ending at \(b\). Write the points as
The \(i\)th subinterval runs from \(x_{i-1}\) to \(x_i\). Its width is the right endpoint minus the left endpoint:
Because the partition points are strictly increasing, each width is positive. A partition is unequal when its subinterval widths are not all the same. The symbol \(\Delta x_i\) refers to the width of the \(i\)th subinterval; the index matters because another subinterval may have a different width.
If \(x\) measures time in minutes and \(r_i\) is measured in litres per minute, then \(\Delta x_i\) is in minutes and \(r_i\Delta x_i\) is in litres. If the rate is negative, the contribution is negative: the sign records the direction of change, while the width remains positive.
Why the Widths Add to the Whole Interval
The individual widths are differences between consecutive partition points. Adding these differences makes every interior point cancel: it appears once as a right endpoint and once as a left endpoint. Only the final endpoint and starting point remain.
This is a useful check on width calculations. It does not say that the widths are equal; it says their total must equal the length of the original interval. The proof also shows why the subtraction order matters: using right endpoint minus left endpoint gives positive widths and the correct total.
Pair Each Width with Its Own Rate
Suppose a rate sample \(r_i\) is given for each interval \([x_{i-1},x_i]\). The corresponding finite contribution is \(r_i\Delta x_i\). Adding the contributions gives the finite accumulation model
This pairing is part of the meaning of the sum: \(r_i\) and \(\Delta x_i\) both refer to the same interval. If the rate is in units of quantity per unit of \(x\), multiplying by a width in units of \(x\) gives a contribution in units of quantity. The calculation accounts for the supplied rate samples; it does not determine how those samples were chosen.
Worked Example: Finding Widths and Contributions
Partition the interval \([0,10]\) at \(x_0=0\), \(x_1=1.5\), \(x_2=4\), and \(x_3=10\). Suppose the rate samples on the three successive subintervals are \(3\), \(2\), and \(-1\) litres per minute. Find the widths and the total finite accumulation.
First calculate each width by subtracting consecutive endpoints:
The widths are unequal. Check their total against the full interval length:
Now match each width with the rate on its own interval:
Therefore the total is \(4.5+5-6=3.5\) litres. Check the products by dividing each contribution by its width: \(4.5/1.5=3\), \(5/2.5=2\), and \(-6/6=-1\) litres per minute, which recovers the given rates. The sign of the last contribution is negative because its rate is negative.
Keep the Interval Pairing Visible
A table can make it harder to lose track of which rate belongs to which width. Keep the interval endpoints, width, rate, and contribution on the same row. The row index then records the pairing.
Worked Example: A Time Partition with Four Unequal Widths
A flow rate is sampled on the intervals determined by the times \(0\), \(2\), \(5\), \(6.5\), and \(10\) minutes. The corresponding rate samples are \(4\), \(1.2\), \(-2\), and \(3\) cubic metres per minute. Compute the finite accumulation from these samples.
Subtract adjacent times to obtain the four widths:
The widths sum to \(2+3+1.5+3.5=10\) minutes, which agrees with the full duration \(10-0=10\) minutes. Pair each rate with the width for its interval:
| Interval (min) | Width (min) | Rate (m³/min) | Contribution (m³) |
|---|---|---|---|
| \([0,2]\) | 2 | 4 | \(4(2)=8\) |
| \([2,5]\) | 3 | 1.2 | \(1.2(3)=3.6\) |
| \([5,6.5]\) | 1.5 | \(-2\) | \((-2)(1.5)=-3\) |
| \([6.5,10]\) | 3.5 | 3 | \(3(3.5)=10.5\) |
Add the contributions: \(8+3.6-3+10.5=19.1\) cubic metres. To check the interval matching, divide each contribution by its row’s width: \(8/2=4\), \(3.6/3=1.2\), \(-3/1.5=-2\), and \(10.5/3.5=3\). Each quotient agrees with the rate sample assigned to that interval.
Changing Units or Endpoint Values
Widths need not be whole numbers. They may be fractions or decimals, and the same subtraction rule applies. Keep the endpoint order clear, retain enough precision during the calculation, and check the sum of the widths before using the rate samples.
Worked Example: Fractional Widths and a Negative Rate
Consider partition points \(x_0=-2\), \(x_1=-1\), \(x_2=1\), and \(x_3=4\). Suppose the rate samples on the successive intervals are \(5\), \(-2\), and \(3\) units per unit of \(x\). Find the widths and the total finite contribution.
Calculate each width as right endpoint minus left endpoint:
Their sum is \(1+2+3=6\), matching the full interval length \(4-(-2)=6\). The paired contributions are
The total is \(5-4+9=10\) units of accumulated quantity. Check by dividing each contribution by its width: \(5/1=5\), \(-4/2=-2\), and \(9/3=3\). These are precisely the given rates. In particular, the negative rate produces a negative contribution even though the width \(2\) is positive.
Common Mistake: Treating Unequal Widths as Equal
A frequent error is to find the full interval length and divide it equally among the subintervals, even when the partition points show otherwise. For example, the partition \(0,2,3,7\) has total length \(7\) and three subintervals. Assigning width \(7/3\) to each would be incorrect: the actual widths are
Suppose the associated rate samples are \(5\), \(-2\), and \(3\). The incorrect equal-width calculation gives
The correction is to pair each rate with the width of its own interval:
The correct widths add to \(2+1+4=7\), as required. The equal-width calculation has substituted widths that do not match the partition, so it does not represent the stated rate-and-interval pairings.
A Reliable Calculation Routine
Write the partition points from the start of the interval to its end, and identify the subinterval associated with each rate sample.
For each interval, compute right endpoint minus left endpoint. This gives its positive width.
Add them and confirm that their total equals the full interval length.
Multiply each rate sample by the width of its own interval, retaining the sign of the rate.
Add the contributions with units, then divide each contribution by its width to confirm that the paired rate is recovered.
The key habit is to keep the index pairing visible: interval \(i\), width \(\Delta x_i\), and rate \(r_i\) belong together. This prevents both width errors and accidental reassignment of a rate to a neighboring interval.
Check Your Understanding
For each question, keep the interval endpoints and their widths explicit before combining any rate samples.
- For partition points \(0\), \(1\), \(4\), and \(6\), calculate all three widths and verify their sum.
- A partition has widths \(2\), \(0.5\), and \(3.5\). What is the length of the full interval?
- On intervals with widths \(2\) and \(5\), the rate samples are \(4\) and \(-1\) units per minute. What are the two signed contributions?
- Why is it incorrect to use the same width for every subinterval when the partition points are unequal?
- If a contribution is \(-6\) units and its interval width is \(3\) minutes, what rate sample was paired with that interval?
