A rate says how fast a quantity changes right now; the total change over an interval is what that rate adds up to. When the rate is constant the sum is a single multiplication and it is exact. When the rate moves, cut the interval into pieces, freeze the rate in each piece, and add rate times width. The result is an exact fraction and an estimate, and the lesson is about keeping those two facts apart.
Core ideas
Three ideas, in the order a reader meets them: the case where nothing needs estimating, the device that handles every other case, and the cost of that device.
Water enters a tank at 3 litres per minute for the 4 minutes from t = 1 to t = 5: the tank gains 3·4 = 12 litres. Nothing else is needed, and the lab agrees: on the constant preset every sum it prints is 12, whatever number of pieces you ask for.
Cut the interval into n pieces of equal width h = (b − a)/n. Inside one piece, pretend the rate is the number it has at the piece's left end. That piece then contributes rate times width, a constant-rate product, and the pieces are added. This is the left sum, written Lₙ.
In the real situation the rate keeps changing inside each piece. If it is climbing, the left-end value is the smallest the rate takes in that piece, so every piece is short and the whole sum is short. The sum is an exact fraction, and it is still only an estimate of the total.
The material
The constant case, the device for everything else, and what the device quietly assumes.
Distance is speed times time, and the same sentence is true of every rate: the change in a quantity is the rate of change times the elapsed time, provided the rate does not change. A tank filling at a steady 3 litres per minute gains 3·4 = 12 litres in 4 minutes. The proviso is the whole difficulty, because most rates worth asking about move.
Let a rate f(t) be given on the interval from a to b. Cut the interval into n pieces of equal width h = (b − a)/n, with left ends a, a + h, a + 2h, …. The left sum is the total of rate times width, using the rate at each left end.
Lₙ = h·(f(a) + f(a + h) + f(a + 2h) + … + f(b − h)). It is exact when f is constant, and an estimate otherwise.
Take the rate 2t on the interval from 0 to 2: a tap that opens slowly, so that t minutes after it opens it is delivering 2t litres per minute. With n = 4 the width is h = 1/2, the left ends are 0, 1/2, 1, 3/2, and the rates there are 0, 1, 2, 3. Each piece contributes its rate times 1/2.
The sum is 3. The lab's exact-total tile prints 4, and the gap is not a rounding: the tap is still opening during every piece, and each piece used the smallest rate it contains. On a piece from 1 to 3/2 the rate climbs from 2 to 3, and the sum charged the whole piece at 2. The missing part of each piece is a small triangle of width 1/2 and height 1, of area 1/4, and four of them make the missing 1.
Where the total 4 comes from is the work of “The Antiderivative and the Fundamental Theorem”, later in this course. For now it is the number the sums are compared with, and the next lessons show them closing in on it. What matters here is the direction of the error: a rising rate and a left sum give a sum that is too low.
A tempting shortcut for 2t on the interval from 0 to 2 is the rate at the end, 4, times the elapsed time, 2, which gives 8. That is twice the lab's total of 4.
The shortcut treats the rate as 4 for the whole two minutes, but it was 4 only at the very end. It was 0 at the start. Freezing the rate at its last value overstates the total by the same kind of error the left sum understates it, in the other direction and by more.
A steeper rate shows the same thing at a larger scale. The speed preset is t² + 1 on the interval from 0 to 3 in n = 3 pieces of width 1. The left ends are 0, 1, 2, the rates there 1, 2, 5, and Lₙ = 1·(1 + 2 + 5) = 8, against an exact total of 12. Three pieces are far too few for a rate that climbs from 1 to 10.
Raising the number of pieces helps, and the ramp shows it plainly: with n = 8 the left sum of 2t is 7/2, and with n = 16 it is 15/4. Both are still below 4, both are exact fractions, and neither is the total. How far the process can be pushed, and what it approaches, are the questions of the next two lessons.
Interactive
Left, right, trapezoid and midpoint sums as fractions
Method
Five moves. The last one is the one that is easy to skip and the one that tells you how far to trust the answer.
Write f(t) and the two ends a and b. Say what the units are: litres per minute times minutes is litres. If the rate is constant, multiply and stop; nothing below is needed.
Pick n, then h = (b − a)/n. For the ramp, (2 − 0)/4 = 1/2. Equal widths keep every later step a single multiplication.
The left ends are a, a + h, …, b − h: n values and not n + 1, because the point b is the left end of no piece. Evaluate f at each.
Add the rates first and multiply by h once, or multiply piece by piece; the result is the same fraction. If you find yourself adding the rates and stopping, you have forgotten the width.
The rate was frozen at the left end of each piece. If it rises across the interval the sum is low; if it falls the sum is high. Say so beside the number, because the number alone looks exact.
A tap opens slowly and delivers 2t litres per minute, t minutes after it opens. Estimate the water collected over the first two minutes with four pieces, using the rate at each left end.
The left sum is 3 and the total is 4; the difference is exactly the four small triangles the frozen rate cut off. The shortcut 8 is further out than the estimate, which is a reason to prefer an honest estimate over a convenient formula.
For a rehearsal, use the speed preset and write out the three left ends and rates yourself. The supplied first move is that the rates are 1, 2, 5. Add them, multiply by the width 1, and then say in one sentence why the answer is below 12.
You should be able to compute the width and the left ends for a given number of pieces, evaluate the rate at each, form the left sum as an exact fraction, and say whether it is below or above the total for a rising or falling rate and why. You should also be able to say, in one sentence, what is wrong with the end rate times the elapsed time.
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