Freezing each piece at its left end gives one estimate; freezing it at its right end gives another. Their difference can be written down before either is computed, because every term but two cancels. When the rate only rises or only falls, the total change is trapped between the two sums, and which sum is the larger depends on the direction of the rate and on nothing else.
Core ideas
Three ideas: a second estimate, the formula for how far apart the two are, and the condition under which the pair brackets the truth.
The same pieces, the same widths, but the rate is read at the right end of each piece: Rₙ = h·(f(a + h) + f(a + 2h) + … + f(b)). The two sums use almost the same list of rates. The left sum uses f(a) and not f(b); the right sum uses f(b) and not f(a).
Subtract the two sums and every interior rate appears once with each sign. What survives is Rₙ − Lₙ = h·(f(b) − f(a)). The gap can be written down from the end values alone, and it halves each time the number of pieces doubles, because h does.
If the rate only rises, each piece's true contribution lies between its left-end estimate and its right-end estimate, so the total lies between Lₙ and Rₙ. If it only falls, the roles swap. The bracket is the first honest statement about the total that needs no further information.
The material
The right sum, the telescoping gap and its proof, and the conditions on the rate that the bracket needs.
The previous lesson froze the rate at the left end of each piece. Freezing it at the right end is equally defensible, and the two give different numbers. Using the same four pieces on the square rate t² over the interval from 0 to 1, the width is 1/4, the left rates are 0, 1/16, 4/16, 9/16, and the right rates are 1/16, 4/16, 9/16, 16/16.
With the pieces of the previous lesson, the right sum uses the rate at each right end: Rₙ = h·(f(a + h) + … + f(b)). The gap is Rₙ − Lₙ.
A rate is rising if f never decreases from a to b, and falling if it never increases. Monotone means one or the other.
For any rate f on the interval from a to b cut into n equal pieces of width h, Rₙ − Lₙ = h·(f(b) − f(a)).
Write the left ends as t₀ = a, t₁, …, tₙ₋₁ and the right ends as t₁, …, tₙ = b. Then Lₙ = h·(f(t₀) + f(t₁) + … + f(tₙ₋₁)) and Rₙ = h·(f(t₁) + … + f(tₙ₋₁) + f(tₙ)).
The terms f(t₁), …, f(tₙ₋₁) appear in both, so they cancel in the difference. What remains is h·(f(tₙ) − f(t₀)), which is h·(f(b) − f(a)). Nothing was assumed about f: this holds for a rate that wanders as much as you like.
□
The proof has no condition on the rate, and the bracket does. Suppose the rate rises. On any one piece the rate is at least its left-end value and at most its right-end value, so the true change over that piece is at least rate-at-left times width and at most rate-at-right times width. Adding over the pieces, Lₙ ≤ total ≤ Rₙ. For a falling rate every inequality reverses.
The right sum is not always the overestimate. The falling preset is 4 − t on the interval from 0 to 2, with four pieces. Its left-end rates are 4, 7/2, 3, 5/2, giving Lₙ = 13/2, and its right-end rates are 7/2, 3, 5/2, 2, giving Rₙ = 11/2. The gap Rₙ − Lₙ is −1, which is (1/2)·(2 − 4), the formula applied without change. The lab's exact total is 6, between the two with the larger sum on the left.
Type t - t^2 as the rate, with the interval from 0 to 1 and one piece. The rate is 0 at both ends, so both sums are 0, and the lab's exact total is 1/6.
The total is outside the pair. The bracket needed the rate to move one way only; this rate rises and then falls, and the two end values say nothing about the middle. The gap formula still holds, and gives 0.
Every number the lab prints for the gap is an exact fraction, and so is the bracket. On the square preset the total 1/3 lies between 7/32 and 15/32, a window of width 1/4. That is a statement the page can prove by arithmetic rather than assert, and doubling the pieces halves the window.
Interactive
Left, right, trapezoid and midpoint sums as fractions
Method
The order matters: decide the direction of the rate before you decide which sum is the lower one.
Rising, falling, or neither. For a polynomial rate, check the ends and ask whether it turns around in between. If it is not monotone, the bracket is not available and the rest is only two estimates.
h·(f(b) − f(a)) from the end values and the width. It is the width of the window the total will sit in, and it is a check on your sums: they must differ by exactly this.
Rates at a, a + h, …, b − h, added and multiplied by h.
Rates at a + h, …, b; or simply Lₙ + h·(f(b) − f(a)), which is the same fraction and a useful independent check.
For a rising rate the left sum is lower and the right sum higher; for a falling rate the reverse. State the bracket as lower ≤ total ≤ higher, and nothing stronger.
The rate t² on the interval from 0 to 1 is rising. Compute both sums with four pieces and check the gap formula.
The two sums differ by 1/4 and the formula said they would, before either sum was computed. The total 1/3 lies between them, and it is much nearer the left sum than the right. Nothing in the bracket tells you that; it is only a window.
For a rehearsal, take the falling preset. The supplied first move is that the rate drops by 2 across the interval. Predict the gap from that, then check it against the lab, and say which sum is the larger and why.
You should be able to compute the left and right sums as exact fractions, predict the gap from the end values of the rate and the width alone, say which sum is the larger for a rising and a falling rate, and state the bracket for a monotone rate without claiming more than it gives.
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