Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.
The problem
In Topic 6.1 every rate graph was made of straight lines or circular arcs, so geometry gave the exact area. Most graphs are curves. There is no formula from geometry for the area under y = x²/2 + 2, for example.
The plan is to approximate. Cut the interval into strips. Replace each strip with a shape whose area you can find (a rectangle or a trapezoid). Add the areas. More, thinner strips usually give a better estimate.
The exact value you are estimating is called a definite integral. Topic 6.3 gives its notation and formal definition. On this page, “the exact area” means that value: the signed area between the graph and the x-axis from a to b.
Partitions and widths
A partition of [a, b] is a list of points
a = x₀ < x₁ < x₂ < … < xₙ = b
It cuts [a, b] into n subintervals. The width of the i-th subinterval is Δxᵢ = xᵢ − xᵢ₋₁.
- In a uniform partition every width is the same: Δx = (b − a)/n.
- In a nonuniform partition the widths differ. This is common with tables, where readings were taken at uneven times.
The four sums
On each subinterval, you choose a height and multiply by the width.
| Sum | Height used on each subinterval | One term |
|---|---|---|
| Left Riemann sum, L | value at the left end | f(xᵢ₋₁) × Δxᵢ |
| Right Riemann sum, R | value at the right end | f(xᵢ) × Δxᵢ |
| Midpoint Riemann sum, M | value at the middle | f((xᵢ₋₁ + xᵢ)/2) × Δxᵢ |
| Trapezoidal sum, T | average of the two ends | ½[f(xᵢ₋₁) + f(xᵢ)] × Δxᵢ |
The first three use rectangles. The trapezoidal sum uses trapezoids, which follow a sloping graph more closely.
The heights can come from a formula, from reading a graph, from a table, or from a description in words. If f is negative somewhere, those terms are negative, just as an area below the axis counted as a decrease in Topic 6.1.
A useful link. For the same partition, the trapezoidal sum is the average of the left and right sums:
T = (L + R)/2
Each trapezoid term is the average of a left term and a right term, so the whole sum is too.
Worked example 1: a function given by a formula
Question. Let f(x) = x²/2 + 2. Approximate the area under f from x = 0 to x = 6 using three subintervals of equal width, with (a) a left sum, (b) a right sum, (c) a trapezoidal sum and (d) a midpoint sum.
Set up. Δx = (6 − 0)/3 = 2. The partition is 0, 2, 4, 6.
| x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| f(x) | 2 | 2.5 | 4 | 6.5 | 10 | 14.5 | 20 |
(a) Left sum. Heights at 0, 2 and 4: L = 2 × (2 + 4 + 10) = 2 × 16 = 32
(b) Right sum. Heights at 2, 4 and 6: R = 2 × (4 + 10 + 20) = 2 × 34 = 68
(c) Trapezoidal sum. T = 2 × [½(2 + 4) + ½(4 + 10) + ½(10 + 20)] = 2 × (3 + 7 + 15) = 50 Check: (L + R)/2 = (32 + 68)/2 = 50.
(d) Midpoint sum. The midpoints are 1, 3 and 5: M = 2 × (2.5 + 6.5 + 14.5) = 2 × 23.5 = 47
How good are they? The exact area is 48 (you will be able to find this yourself later in this unit). The left sum is 16 too small and the right sum 20 too big. The trapezoidal sum is 2 too big and the midpoint sum only 1 too small. The figures below show why.
Over or under? Reading the behaviour of f
You can often tell whether a sum is too big or too small without knowing the exact value. You need to know how f behaves on the whole interval [a, b].
| What you know about f on [a, b] | Left sum L | Right sum R | Midpoint M | Trapezoidal T |
|---|---|---|---|---|
| increasing | under | over | (use concavity) | (use concavity) |
| decreasing | over | under | (use concavity) | (use concavity) |
| concave up | (use monotonicity) | (use monotonicity) | under | over |
| concave down | (use monotonicity) | (use monotonicity) | over | under |
Why left and right depend on increasing or decreasing. If f is increasing, the left end of each subinterval is the lowest point on it. A rectangle of that height fits under the graph, so L is too small. The right end is the highest point, so R is too big. For a decreasing f, swap them.
Why the trapezoid depends on concavity. A concave-up graph bends upward, so it lies below each chord joining two of its points. The trapezoid tops are those chords. They sit above the curve, so T is too big. For concave down the chords sit below the curve, so T is too small.
Why the midpoint is the other way round. Tilt the top of a midpoint rectangle until it lies along the tangent line at the midpoint. The tilt does not change the area, because the strip gained on one side equals the strip lost on the other. A concave-up graph lies above its tangent lines, so the tilted top, and therefore M, is too small. For concave down, M is too big.
How to justify on paper. Name the property and connect it to the sum: “f is decreasing on [0, 8], so on each subinterval the left end gives the largest value. The left Riemann sum is therefore an overestimate.” A claim with no reason earns little credit.
Two cautions.
- A table of values alone does not tell you how f behaves between the readings. You need a statement such as “f is increasing” or information about f′ or f″.
- If f is increasing on part of [a, b] and decreasing on another part, the rule does not apply to the whole sum. The sum might be too big or too small.
Worked example 2: a table with uneven gaps
Question. A file is downloaded over 40 seconds. The download speed S(t), in megabits per second (Mb/s), is measured at a few times t, in seconds. S is increasing on 0 ≤ t ≤ 40.
| t (s) | 0 | 10 | 15 | 25 | 40 |
|---|---|---|---|---|---|
| S(t) (Mb/s) | 12 | 18 | 26 | 30 | 34 |
(a) Use a right Riemann sum with the four subintervals in the table to estimate the amount of data downloaded from t = 0 to t = 40. (b) Is this an overestimate or an underestimate? Explain. (c) Find the trapezoidal sum with the same subintervals.
Set up. The widths are 10, 5, 10 and 15 seconds. They are not equal, so each term needs its own width.
(a) R = 10 × 18 + 5 × 26 + 10 × 30 + 15 × 34 = 180 + 130 + 300 + 510 = 1,120 megabits.
Units: (Mb/s) × s = Mb. About 1,120 megabits were downloaded in the first 40 seconds. (S is a rate, so the sum estimates the accumulated change in data, as in Topic 6.1.)
(b) S is increasing, so on each subinterval the right end gives the largest speed. Each rectangle is at least as tall as the graph across its subinterval. The right sum is an overestimate.
(c) The left sum is 10 × 12 + 5 × 18 + 10 × 26 + 15 × 30 = 120 + 90 + 260 + 450 = 920. Then
T = (920 + 1,120)/2 = 1,020 megabits.
Or directly: 10 × ½(12 + 18) + 5 × ½(18 + 26) + 10 × ½(26 + 30) + 15 × ½(30 + 34) = 150 + 110 + 280 + 480 = 1,020.
Can you say whether T is over or under? Not from this information. The question tells you S is increasing, but says nothing about concavity.
Midpoint? A midpoint sum on these subintervals would need S at t = 5, 12.5, 20 and 32.5. The table does not give those, so a midpoint sum is not possible here.
Worked example 3: a function given by a graph and in words
Question. A cyclist’s speed v(t), in metres per second, is shown in Figure 3 for 0 ≤ t ≤ 12 seconds. The graph passes through the marked points. You are also told, in words, that the graph of v is concave down on the whole interval.
(a) Estimate the distance travelled with a midpoint sum on two subintervals of equal width. (b) Estimate it with a trapezoidal sum on four subintervals of equal width. (c) Say whether each estimate is too big or too small.
(a) The subintervals are [0, 6] and [6, 12]. Their midpoints are t = 3 and t = 9, where the graph reads 7 and 7. M = 6 × 7 + 6 × 7 = 84 m.
(b) Width 3, with heights 4, 7, 8, 7, 4 read from the graph: T = 3 × [½(4) + 7 + 8 + 7 + ½(4)] = 3 × 26 = 78 m.
(c) The words tell you v is concave down. So T is an underestimate and M is an overestimate, and the distance lies between 78 m and 84 m. (It is in fact 80 m.) Without the sentence about concavity, the five dots alone would not settle this.
Using technology
With many subintervals, the arithmetic is long, so use a calculator list or a spreadsheet. The thinking stays the same.
- Write the partition and the widths.
- Decide which height each term uses.
- Write the sum in full (or with the first few terms and the last term) before giving its value.
On a calculator-active exam question, a bare number with no set-up shows no method. Writing “10 × 18 + 5 × 26 + 10 × 30 + 15 × 34” first shows your method even if you then make a keying error.
Common misconceptions
- Using one Δx for a nonuniform table. Every term needs its own width.
- Counting the wrong number of terms. n subintervals give n terms. A left sum never uses the last value; a right sum never uses the first.
- “Midpoint” from a table. The midpoint of a subinterval is a point in the middle of the x-interval, not the average of the two table values.
- Mixing up which property decides what. Increasing/decreasing decides left and right. Concavity decides midpoint and trapezoid.
- Claiming over or under from a table alone. You need information about f between the data points.
- Forgetting the ½ in a trapezoid term. Without it you get L + R, double the trapezoidal sum.
- Giving a sum with no units or context. The sum estimates an accumulated change, so it carries units of (rate) × (input).
Where this leads
Topic 6.3 writes these sums with sigma (Σ) notation, and defines the definite integral as the limit of a Riemann sum as the widths shrink to zero. Later in the unit, the Fundamental Theorem of Calculus gives exact values, which you can compare with your estimates. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.
Try the practice questions now, then use the revision notes and the checklist to consolidate.
Quick questions
Is this page for Calculus AB or Calculus BC?
Both. Topic 6.2 is common content, so the same page serves AB and BC students.
Do the subintervals have to be the same width?
No. Tables often have uneven gaps. Use the actual width of each subinterval in each term.
Can I always use a midpoint sum with a table?
Only if the table gives the function value at the middle of each subinterval you choose. Otherwise use left, right or trapezoidal sums.