From Rate Samples to an Accumulation Estimate
A rate describes how quickly a quantity changes. To estimate the total change over an interval, divide the interval into smaller pieces, estimate the change on each piece, and add those estimates. The basic idea is rate multiplied by duration: a rate sample represents the rate over a piece, and the piece’s width represents its duration.
A rectangle estimate implements this idea geometrically: each rectangle has a height given by a rate sample and a width given by a subinterval’s width. In a left-endpoint sum, the rate for each subinterval is sampled at that subinterval’s left end. The sample and width must be paired with the same subinterval.
Definition for Any Partition
Let \(f\) be a rate function on an interval \([a,b]\). A partition of this interval consists of increasing points \(x_0,x_1,\ldots,x_n\) with \(x_0=a\) and \(x_n=b\). Its \(i\)th subinterval is \([x_{i-1},x_i]\), and its width is \(x_i-x_{i-1}\).
The index \(i\) labels the subinterval, not the sample point. The \(i\)th subinterval begins at \(x_{i-1}\), so the sample uses \(x_{i-1}\), while the width uses both endpoints, \(x_i-x_{i-1}\). A partition need not have equal widths; the formula uses the actual width of each subinterval.
If \(f\) has units of “quantity per unit of input” and \(x\) has units of the input, each product \(f(x_{i-1})(x_i-x_{i-1})\) has units of the accumulated quantity. For example, a rate in liters per minute multiplied by a time width in minutes gives an amount in liters.
Why the Sum Estimates Accumulation
On one subinterval, suppose we use the starting rate \(f(x_{i-1})\) as a stand-in for the rate throughout that piece. The estimated change on that subinterval is the stand-in rate multiplied by the width. Adding the estimated changes from all the subintervals gives the left-endpoint sum.
and therefore
This explains the construction: the sum adds the estimated contributions of the separate pieces. If the rate varies within a subinterval, the starting sample may not represent the whole piece exactly, so the result is an estimate rather than automatically the exact change. More subintervals provide more samples, but this construction by itself does not guarantee a particular error size.
A Reliable Construction Process
Write the endpoints of every subinterval, from the beginning of the full interval to its end.
For \([x_{i-1},x_i]\), use \(x_{i-1}\) as the input to the rate function or table.
Calculate \(x_i-x_{i-1}\) for that same subinterval.
Multiply each left-endpoint sample by its matching width, then add all the products.
Use the context to state what the estimate represents and give its units. If the rate can be negative, retain the signs of the products.
Worked Examples
Worked Example: Unequal Time Intervals and Velocity
A vehicle’s velocity samples at times \(t=0\), \(t=2\), and \(t=5\) seconds are \(4\), \(3\), and \(5\) meters per second, respectively. Use the partition \(0<2<5<6\) to estimate displacement with a left-endpoint sum.
The subintervals are \([0,2]\), \([2,5]\), and \([5,6]\). Their widths are \(2-0=2\) seconds, \(5-2=3\) seconds, and \(6-5=1\) second. Their left endpoints are \(0\), \(2\), and \(5\), where the given velocity samples are \(4\), \(3\), and \(5\) meters per second.
The three contributions are \(8\) meters, \(9\) meters, and \(5\) meters. Thus
Check: The products \(4\cdot2=8\), \(3\cdot3=9\), and \(5\cdot1=5\) each use the rate at the start of the corresponding interval. Their sum is \(8+9+5=22\), and velocity times seconds has units of meters. The estimated displacement is \(22\) meters.
Worked Example: Flow Rates from a Table
A tank fills at the rates shown below. Use the left-endpoint sum on the partition \(0<1<4<6\) minutes to estimate the volume added over those six minutes.
| Time (minutes) | Flow rate (liters per minute) |
|---|---|
| 0 | 12 |
| 1 | 9 |
| 4 | 15 |
The subintervals are \([0,1]\), \([1,4]\), and \([4,6]\), with widths \(1\), \(3\), and \(2\) minutes. The left-endpoint rates are \(12\), \(9\), and \(15\) liters per minute, in that order. Therefore,
The contributions are \(12\) liters, \(27\) liters, and \(30\) liters. Adding gives
Check: Each rate comes from the table row at the start of its interval: time \(0\) for width \(1\), time \(1\) for width \(3\), and time \(4\) for width \(2\). The products sum to \(69\), and liters per minute multiplied by minutes gives liters. The estimated volume added is \(69\) liters.
Worked Example: A Rate Formula on Unequal Subintervals
Suppose a quantity changes at rate \(f(t)=t^2+1\) units per second. Estimate its change from \(t=1\) to \(t=5\) using the partition \(1<2<4<5\).
The widths are \(2-1=1\), \(4-2=2\), and \(5-4=1\) second. Evaluate the rate at each left endpoint:
Now multiply each value by its matching width:
The rate’s units are units per second, so the estimated change is \(29\) units. Check: the three products are \(2\), \(10\), and \(17\), and their sum is \(29\). The sample inputs are precisely the left endpoints \(1\), \(2\), and \(4\); the widths are \(1\), \(2\), and \(1\). Thus the terms match the stated partition.
Signed Rates and Net Change
A rate can be negative as well as positive. In that case, its product with a positive interval width is negative, so that subinterval contributes a negative amount to the estimated net change. Do not replace negative samples with their absolute values when the goal is net change.
For instance, suppose the left-endpoint rate samples on three intervals of widths \(1\), \(2\), and \(1\) are \(-2\), \(1\), and \(3\) units per second. The sum is
in units of the accumulated quantity. The negative first contribution remains negative; the later positive contributions partly offset it. This is a signed estimate of net change, not a sum of the magnitudes of all changes.
Common Indexing and Pairing Errors
A left-endpoint sum has two pieces for each interval: the sample at the start and the width of that same interval. It is easy to calculate the correct numbers but attach a width to the wrong rate.
The correct widths in order are \(2-0=2\), \(5-2=3\), and \(6-5=1\). Keep each rate and width together by writing one term per interval:
A useful check is to make a short row for every interval showing its left endpoint, its rate sample, its width, and its product. If the listed widths do not add to the full interval length, there is an error: here \(2+3+1=6\), which matches the interval from \(0\) to \(6\).
What the Construction Gives You
A left-endpoint sum is a systematic way to turn sampled rates into an estimate of accumulated change. Its essential structure is simple, but accurate use depends on attending to the partition: identify each subinterval, sample at its left end, use that subinterval’s own width, and preserve the sign of the rate.
The result should be interpreted with care. The sum records the contributions estimated from the chosen starting samples. When a rate changes across a subinterval, the sampled value need not equal the rate at every point there. Consequently, the sum is generally an estimate, and its usefulness depends on how well the samples represent the rate over their intervals.
Check Your Understanding
For each question, keep the sample locations and widths associated with the same subinterval.
- For the partition \(2<3<7\), what is the left endpoint of each subinterval, and what are the two widths?
- A rate has left-endpoint samples \(6\) and \(4\) units per minute on intervals of widths \(2\) and \(3\) minutes. Write the left-endpoint sum and state its units.
- For \(f(x)=2x+1\) and the partition \(0<1<3\), write and evaluate the left-endpoint sum.
- Why must a negative rate sample produce a negative term when estimating net change over a positive-width interval?
- In a left-endpoint sum, which endpoint supplies the rate sample for \([x_{i-1},x_i]\), and how is the width calculated?
