Why the Midpoint Can Be Exact
A rectangle is usually an estimate of the area under a rate graph. For a rate that changes in a straight line, however, the midpoint rectangle does more than estimate: on each interval, its signed area is exactly the signed area between the graph and the time axis.
The reason is a symmetry in linear change. Over an interval, the rate at the midpoint is the average of its values at the two endpoints. The exact signed area for a straight-line graph is also the interval width multiplied by that endpoint average. These are two expressions for the same quantity.
This result concerns signed accumulation: area above the time axis counts positively and area below it counts negatively. It applies whether the rate remains positive, remains negative, or crosses zero.
Exact Signed Area for a Linear Rate
Let \(r(t)=mt+c\) be a linear rate on an interval \([u,v]\), where \(u<v\). Write \(\Delta t=v-u\) for the interval width. The signed area is the area above the axis minus the area below it, with each part measured geometrically. For a straight-line graph, it can be computed using the endpoint rates:
If both endpoint rates are positive, the graph forms a trapezoid above the axis. If both are negative, the trapezoid is below the axis and its signed area is negative. If the graph crosses the axis, the formula still gives the signed area: the part below the axis is subtracted from the part above it.
For instance, suppose the endpoint rates are \(p\) and \(-q\), with \(p,q>0\). Since the graph is linear, it reaches zero after a fraction \(p/(p+q)\) of the interval width. The positive triangle has base \(\Delta t\cdot p/(p+q)\) and height \(p\); the negative triangle has base \(\Delta t\cdot q/(p+q)\) and height \(q\). Their signed difference is
Because the endpoint rates are \(p\) and \(-q\), this is exactly \(\frac{\Delta t}{2}(p+(-q))\), as in the signed trapezoid formula.
Why the Midpoint Rectangle Is Exact
The midpoint of \([u,v]\) is \(m=(u+v)/2\). For a linear rate, its value there is the average of the endpoint rates. To see this directly, use \(r(t)=mt+c\) with slope \(m\) replaced, to avoid confusing slope and midpoint, by \(r(t)=kt+c\):
A midpoint rectangle has width \(v-u\) and height \(r((u+v)/2)\). Its signed area is width times height. Substituting the midpoint identity gives the endpoint formula for exact signed area.
This argument proves the result for any linear rate and any interval of positive width. The rectangle’s height is allowed to be negative; in that case, its signed area is negative as well.
Worked Example: One Interval with a Positive Rate
Worked Example: One Interval with a Positive Rate
A machine’s production rate is \(r(t)=3t+2\) units per hour, where \(t\) is measured in hours. Find the exact signed accumulation from \(t=0\) to \(t=4\) using the midpoint rectangle, then check using endpoint rates.
The interval width is \(4-0=4\) hours. Its midpoint is
The rate at the midpoint is
Therefore, the midpoint rectangle gives
For a check, the endpoint rates are \(r(0)=3(0)+2=2\) units per hour and \(r(4)=3(4)+2=14\) units per hour. The signed trapezoid area is
The two calculations agree, so the exact signed accumulation is \(32\) units.
Applying the Result Across a Partition
Divide a time interval into subintervals and use one midpoint rectangle on each. If the rate is linear on the whole interval, it is linear on every subinterval too. The midpoint result therefore applies separately on every piece, whether the subinterval widths are equal or unequal.
Record each pair of endpoints and calculate each width.
For endpoints \(u\) and \(v\), use the sample time \((u+v)/2\), then evaluate the rate there.
Multiply each midpoint rate by its subinterval width, keeping negative contributions negative.
For each piece, use its width times the average of its endpoint rates. The totals must agree.
Because each rectangle is exact on its own subinterval, adding the rectangle areas gives the exact signed area over the entire partition. Equal widths are not required.
Worked Example: Unequal Subintervals
Worked Example: Unequal Subintervals
A flow rate is \(r(t)=10-2t\) liters per minute on \(0\leq t\leq5\). Use the unequal subintervals \([0,1]\), \([1,4]\), and \([4,5]\) to find the exact signed accumulation by midpoint rectangles, and check the result on each subinterval.
The widths are \(1-0=1\), \(4-1=3\), and \(5-4=1\) minute. The midpoints are \((0+1)/2=0.5\), \((1+4)/2=2.5\), and \((4+5)/2=4.5\) minutes. The corresponding midpoint rates are
in liters per minute. Thus the three midpoint rectangle areas are
liters. Their sum is \(9+15+1=25\) liters.
Check each result using the endpoint rates. On \([0,1]\), the endpoint rates are \(r(0)=10\) and \(r(1)=8\), so the signed area is \(1(10+8)/2=9\) liters. On \([1,4]\), they are \(r(1)=8\) and \(r(4)=2\), so the signed area is \(3(8+2)/2=15\) liters. On \([4,5]\), they are \(r(4)=2\) and \(r(5)=0\), so the signed area is \(1(2+0)/2=1\) liter. These add to \(9+15+1=25\) liters, confirming the midpoint total.
As a further check, the endpoint rates on the full interval are \(10\) and \(0\) liters per minute. The signed trapezoid area is \(5(10+0)/2=25\) liters, the same total.
Worked Example: A Rate That Changes Sign
Worked Example: A Rate That Changes Sign
Let \(r(t)=2t-3\) units per minute on \([0,3]\). Find the exact signed accumulation using the midpoint rectangle. Then distinguish it from the total geometric area between the graph and the time axis.
The interval width is \(3\) minutes and its midpoint is \((0+3)/2=1.5\) minutes. The midpoint rate is
The midpoint rectangle therefore has signed area \(3(0)=0\) units. The endpoint rates are \(r(0)=-3\) and \(r(3)=3\), so the exact signed trapezoid area is
For the geometric area without signs, the graph crosses zero at \(2t-3=0\), or \(t=1.5\). Each side is a triangle with base \(1.5\) minutes and height \(3\) units per minute. Each has area \(\frac12(1.5)(3)=2.25\) units, so the total geometric area is \(2.25+2.25=4.5\) units. The signed accumulation is zero because the positive and negative contributions cancel; the total geometric area is not zero.
A Common Mistake: Confusing Net and Total Area
In the sign-changing example, it would be incorrect to say, “The midpoint rate is zero, so there is no area.” The calculation \(3\cdot0=0\) gives the exact signed area, not the total amount of area on both sides of the axis. The correction is to identify which quantity is being asked for. For signed accumulation, keep the negative contribution and add it to the positive one. For total geometric area, count both pieces positively. These are different quantities when a rate changes sign.
Another useful check is to compare the midpoint rectangle with the endpoint average on each interval. For a linear rate they must match exactly. If they do not, recheck the midpoint, interval width, or rate evaluation.
Why the Result Matters
Midpoint rectangles are commonly used as estimates when a rate is curved. A linear rate is a particularly transparent case: it shows why the sample location matters. The midpoint balances the linear change across the interval, so the rectangular height equals the average endpoint height. When a rate is linear on every subinterval, midpoint rectangles give exact signed accumulation on any chosen partition, including one with unequal widths.
Check Your Understanding
For each question, keep track of whether it asks for signed accumulation or total geometric area.
- For a linear rate on \([u,v]\), write the formula for the midpoint time and the exact signed area.
- A linear rate has endpoint values \(4\) and \(10\) on an interval of width \(3\). What is the midpoint rate, and what is the exact signed area?
- Why does the midpoint rate equal the average of the two endpoint rates for a linear function?
- Does the midpoint rectangle result require equal-width subintervals? Explain.
- A linear rate crosses the time axis and has midpoint rate zero. What can be concluded about its signed area, and what cannot be concluded about its total geometric area?
