Why Rectangle-Sum Setups Need Checking
A rectangle estimate combines two different kinds of information: the width of each time interval and a rate used as the rectangle’s height. A setup can look plausible while pairing the wrong width with a rate, leaving out part of the time range, or changing a negative contribution into a positive one.
These errors matter because the sum is intended to estimate an accumulated quantity. If a rate is measured in litres per minute, multiplying it by a time width in minutes gives an amount in litres. A wrong width, height, or sign changes what the sum represents.
For a rectangle corresponding to an interval from \(t_{i-1}\) to \(t_i\), its width is \(t_i-t_{i-1}\). If the height is the rate sampled at a chosen time \(c_i\) in that interval, the rectangle’s signed contribution is the rate at \(c_i\) multiplied by the width.
The sample point might be a left endpoint, a right endpoint, or a midpoint. Those choices can produce different estimates. When checking a setup, use the sample choice the problem specifies; do not silently replace it with another choice.
A Reliable Audit
Check the setup in a fixed order. First identify the full time interval. Then list the smaller intervals and calculate each width from its endpoints. Next identify the sample point for each height. Finally check the sign, units, and arithmetic of every contribution.
Confirm that the rectangles cover the entire time range in the question, without gaps or unintended overlaps.
For an interval from \(t_{i-1}\) to \(t_i\), use \(t_i-t_{i-1}\). Unequal intervals generally have unequal widths.
Read the rate at the sample point specified for that rectangle. Keep the interval and its sample point paired.
A negative rate gives a negative contribution. Rate units multiplied by time units should give the units of the accumulated quantity.
Why Width Times Height Is the Contribution
The basic calculation comes from the meaning of a constant rate. If a quantity changes at a constant rate \(r\) over a time interval of length \(\Delta t\), its change over that interval is rate multiplied by elapsed time. Therefore the change is \(r\Delta t\). A rectangle uses a sampled rate as its height and the interval length as its width, so its signed contribution is \(f(c)\Delta t\).
This explains the structure of a rectangle sum: add the contributions from the intervals being considered. It does not mean that a sampled rate is necessarily constant throughout its interval. The rectangle is an estimate based on the chosen rate sample. The setup still has to use the correct width and sample point for each rectangle.
Worked Example: Unequal Widths Must Stay With Their Own Heights
Worked Example: Unequal Widths Must Stay With Their Own Heights
Suppose the time interval from \(t=0\) to \(t=4\) is divided into \([0,1]\), \([1,3]\), and \([3,4]\). Use right endpoints, with rates \(f(1)=2\), \(f(3)=5\), and \(f(4)=4\), measured in litres per minute. Set up and calculate the estimate.
The widths are \(1-0=1\), \(3-1=2\), and \(4-3=1\) minutes. Because right endpoints are specified, the heights are \(2\), \(5\), and \(4\) litres per minute, respectively. Pair each height with its own interval width:
Check: the intervals run consecutively from \(0\) to \(4\), and their widths add to \(1+2+1=4\) minutes. The units in each product are litres per minute times minutes, so each contribution and the total are in litres. The middle rectangle contributes \(10\) litres because its interval is two minutes wide.
A common width error is to use \(1\) minute for all three rectangles just because there are three rectangles. That would give \(2+5+4=11\) litres. It is wrong: the middle interval is \([1,3]\), whose width is \(2\), not \(1\). The corrected middle term is \(5(2)=10\), giving \(16\) litres in total.
Worked Example: Use the Specified Endpoint
Worked Example: Use the Specified Endpoint
On the interval from \(0\) to \(3\), divide time into \([0,1]\), \([1,2]\), and \([2,3]\). The rate values are \(f(0)=1\), \(f(1)=4\), \(f(2)=2\), and \(f(3)=5\). Find the estimate using right endpoints, then diagnose the setup \(1+4+2\).
Each interval has width \(1\). For a right-endpoint estimate, the sample points are \(1\), \(2\), and \(3\), so the heights are \(4\), \(2\), and \(5\). Thus:
The proposed setup \(1+4+2\) uses the values at \(0\), \(1\), and \(2\). Those are the left endpoints, not the right endpoints. It would be a left-endpoint estimate, with value \(1+4+2=7\), rather than the requested right-endpoint estimate.
Check: the right-endpoint estimate uses one endpoint from each interval, in order: \(1\) for \([0,1]\), \(2\) for \([1,2]\), and \(3\) for \([2,3]\). The calculated total is \(4+2+5=11\). The left-endpoint calculation is \(1+4+2=7\), so the two choices have not been confused.
Neither endpoint choice is automatically the correct one for every problem. The correct choice is the one requested or defined by the setup. If the question specifies midpoints, the height must instead use the rate at each midpoint. An endpoint-based sum cannot be labelled a midpoint estimate simply because it has the same number of rectangles.
Worked Example: Check That No Interval Was Omitted
Worked Example: Check That No Interval Was Omitted
Estimate accumulation on the full interval from \(t=0\) to \(t=4\) using left endpoints on \([0,1]\), \([1,2]\), and \([2,4]\). Suppose the sampled rates are \(f(0)=2\), \(f(1)=3\), and \(f(2)=-1\), in units per minute.
The three widths are \(1-0=1\), \(2-1=1\), and \(4-2=2\) minutes. The left-endpoint samples match the rates given. The contributions are:
If a proposed setup contains only \(2(1)+3(1)=5\), it has omitted the interval \([2,4]\). That interval contributes \((-1)(2)=-2\), so the corrected total is \(5-2=3\).
Check: the listed intervals cover \([0,4]\), because \([0,1]\), \([1,2]\), and \([2,4]\) meet in sequence, and their widths sum to \(1+1+2=4\). The final interval is two minutes long and its negative rate contributes \(-2\). Including all three terms gives \(3\).
One useful coverage check is to write interval endpoints in order and confirm that the first interval starts at the required beginning, the last ends at the required finish, and each interval starts where the previous one ends. Checking only the sum of the widths can miss a gap if an unintended overlap compensates for it.
Worked Example: Preserve Negative Contributions
Worked Example: Preserve Negative Contributions
Over four consecutive one-minute intervals, sampled rates are \(3\), \(2\), \(-1\), and \(-3\) units per minute. Calculate the signed rectangle sum and compare it with a calculation that incorrectly treats all heights as positive.
Each width is \(1\) minute. The signed contributions are \(3(1)=3\), \(2(1)=2\), \((-1)(1)=-1\), and \((-3)(1)=-3\). Therefore:
The incorrect calculation that replaces the negative heights by their positive magnitudes is \(3+2+1+3=9\). It does not give the signed accumulation. A negative rate means the quantity is changing in the opposite direction, so its contribution must reduce the signed total.
Check: the positive contributions sum to \(3+2=5\), while the negative contributions sum to \(-1-3=-4\). Combining them gives \(5+(-4)=1\), agreeing with the rectangle-by-rectangle calculation.
A Common Mistake: Fixing One Error by Creating Another
When a total seems unexpectedly large or small, it can be tempting to change signs, widths, or sample points until the answer looks more plausible. That is not a reliable check. Each term should be tested against the interval and sampling rule that produced it.
For example, in the unequal-width calculation, changing the \(5(2)\) contribution to \(5(1)\) because all rectangles are assumed to have the same width does not repair the setup. It discards the actual two-minute interval. Similarly, turning a negative sampled rate into a positive height changes a signed estimate into a different quantity.
Why These Checks Matter
Rectangle estimates are often built from rates recorded over time, such as flow, speed, or production rate. The accumulated amount depends on how long each rate applies, which is why widths cannot be treated as decorative labels. Sample choices also matter: a left-endpoint rate and a right-endpoint rate need not be equal.
These checks make the meaning of an estimate easier to interpret. If time is measured in hours and the rate in kilometres per hour, each rectangle contributes kilometres. If the sum instead has units of kilometres per hour, a time width has probably been omitted. Units do not prove that every term is correct, but they can reveal an incomplete or mismatched product.
Before accepting a result, check the setup as well as the arithmetic. A correct addition of incorrectly constructed terms is still an incorrect estimate.
Check Your Understanding
For each question, identify what must be checked before accepting the rectangle-sum estimate.
- An interval is split into \([0,2]\) and \([2,5]\). What are the two widths, and why would using width \(2\) for both rectangles be an error?
- A right-endpoint estimate is requested on \([1,2]\) and \([2,4]\). Which sample points supply the heights?
- A setup covers \([0,1]\) and \([2,3]\), while the full interval is \([0,3]\). What part is missing?
- One sampled rate is negative. What sign should its rectangle contribution have, and what does that sign mean for the signed total?
- If a rate is measured in litres per minute and a width in minutes, what are the units of their product?
