AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/6/6-2)
Unit 6 · Topic 6.2
6.2 Approximating Areas with Riemann Sums
When you can't find an area exactly, you can estimate it with rectangles or trapezoids. This topic covers left, right and midpoint Riemann sums and the trapezoidal sum, how to use them on tables with uneven spacing, and how to tell whether an estimate is too high or too low.
Key terms
- left Riemann sum
- right Riemann sum
- midpoint Riemann sum
- trapezoidal sum
- overestimate
- underestimate
The idea: add up rectangles
A Riemann sum estimates the area under a curve on [a, b]. Split [a, b] into smaller subintervals. On each one, build a rectangle whose width is the subinterval's length and whose height is the function's value at one chosen point. Then add the rectangle areas.
Which point sets the height gives the sum its name. A left Riemann sum uses each subinterval's left endpoint. A right sum uses the right endpoint. A midpoint sum uses the middle of each subinterval. A trapezoidal sum doesn't use rectangles at all; it connects the endpoints with a straight segment, which makes a trapezoid with area ½ × (left height + right height) × width.
Formulas you'll actually use
With n equal subintervals of width Δx = (b − a)/n and endpoints x₀ = a, x₁, …, xₙ = b:
- Left sum: Δx [f(x₀) + f(x₁) + … + f(xₙ₋₁)]
- Right sum: Δx [f(x₁) + f(x₂) + … + f(xₙ)]
- Midpoint sum: Δx [f(m₁) + f(m₂) + … + f(mₙ)], where each m is the middle of a subinterval
- Trapezoidal sum: (Δx/2) [f(x₀) + 2f(x₁) + 2f(x₂) + … + 2f(xₙ₋₁) + f(xₙ)]
- The trapezoidal sum always equals the average of the left and right sums.
Tables with uneven widths
Exam tables often have uneven spacing, like t = 0, 2, 5, 9, 10. Then there's no single Δx. Work one subinterval at a time: width × height for rectangles, or width × average of the two heights for trapezoids, and add. Write out every product so a reader can follow your setup.
A midpoint sum from a table needs the function's value at the middle of each subinterval. If the table gives values at 0, 2, 4, 6 and 8, a midpoint sum with two subintervals uses [0, 4] and [4, 8], with heights f(2) and f(6) and width 4 each. You can't build a midpoint sum on [0, 2] from that table, because f(1) isn't given.
Overestimate or underestimate?
For left and right sums, look at whether f is increasing or decreasing. For midpoint and trapezoidal sums, look at concavity. These rules hold when f behaves the same way across the whole interval.
| If f on [a, b] is… | Left sum | Right sum | Midpoint sum | Trapezoidal sum |
|---|---|---|---|---|
| increasing | under | over | — | — |
| decreasing | over | under | — | — |
| concave up | — | — | under | over |
| concave down | — | — | over | under |
Why the rules work
If f is increasing, each left endpoint is the lowest point on its subinterval, so every left rectangle sits under the curve: an underestimate. A trapezoid's top edge is a chord. On a concave up curve, chords lie above the curve, so trapezoids overestimate. The midpoint rectangle has the same area as a trapezoid built from the tangent line at the midpoint, and on a concave up curve tangent lines lie below the curve, so the midpoint sum underestimates. Reason from the picture and you won't need to memorize the table.
You're expected to compute these sums by hand and also with a calculator, and to explain how an estimate compares with the true value.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Left, right and trapezoidal sums from an uneven table
Water flows into a pond at rate R(t) gallons per minute. Selected values: t = 0, 2, 5, 9, 10 minutes; R(t) = 4, 7, 8, 5, 3. Estimate ∫₀¹⁰ R(t) dt using (a) a left Riemann sum, (b) a right Riemann sum and (c) a trapezoidal sum, each with the four subintervals in the table. What does the integral mean?
Show the solutionHide the solution
- Step 1: Widths of the subintervals: 2, 3, 4 and 1.
- Step 2: (a) Left sum uses the left heights 4, 7, 8, 5: 4(2) + 7(3) + 8(4) + 5(1) = 8 + 21 + 32 + 5 = 66.
- Step 3: (b) Right sum uses the right heights 7, 8, 5, 3: 7(2) + 8(3) + 5(4) + 3(1) = 14 + 24 + 20 + 3 = 61.
- Step 4: (c) Trapezoidal sum: ½(4 + 7)(2) + ½(7 + 8)(3) + ½(8 + 5)(4) + ½(5 + 3)(1) = 11 + 22.5 + 26 + 4 = 63.5.
- Step 5: Check: the average of 66 and 61 is 63.5, which matches.
- Step 6: Meaning: gallons per minute × minutes = gallons, so the integral is the total gallons that flowed in from t = 0 to t = 10.
Answer: Left ≈ 66, right ≈ 61, trapezoidal ≈ 63.5 gallons; the integral is the total water (in gallons) that entered the pond during the first 10 minutes.
- Example 2
Over or under? Use the shape of f
Let f(x) = x² + 1 on [0, 4], with 4 equal subintervals. Find the left, right, midpoint and trapezoidal sums, and say which are overestimates of ∫₀⁴ f(x) dx = 76/3 ≈ 25.33.
Show the solutionHide the solution
- Step 1: Δx = 1. Values: f(0) = 1, f(1) = 2, f(2) = 5, f(3) = 10, f(4) = 17.
- Step 2: Left: 1(1 + 2 + 5 + 10) = 18. Right: 1(2 + 5 + 10 + 17) = 34.
- Step 3: Midpoint uses x = 0.5, 1.5, 2.5, 3.5: f = 1.25, 3.25, 7.25, 13.25, so the sum is 25.
- Step 4: Trapezoidal: (1/2)(1 + 2·2 + 2·5 + 2·10 + 17) = (1/2)(52) = 26.
- Step 5: f is increasing on [0, 4], so left (18) is under and right (34) is over.
- Step 6: f″(x) = 2 > 0, so f is concave up: midpoint (25) is under and trapezoidal (26) is over. Both match the exact value 25.33.
Answer: Left 18 (under), right 34 (over), midpoint 25 (under), trapezoidal 26 (over).
Common mistakes
- Using a single Δx when the table's intervals aren't equal. Multiply each height by its own width.
- Including the last value in a left sum or the first value in a right sum. A left sum with n subintervals uses n heights, starting at x₀ and stopping before xₙ.
- Judging a midpoint or trapezoidal estimate by whether f is increasing. Those two depend on concavity.
- Claiming an estimate is an over- or underestimate when the table alone doesn't tell you how f behaves between the points. Only make the claim if you know f is monotonic (always increasing or always decreasing) or know its concavity.
On the exam
- Riemann sums from a table show up very often on free-response questions. Show the products, like 4(2) + 7(3) + …, not just the final number.
- If asked whether an estimate is too high or too low, give the reason: for example, “R is decreasing on the interval, so the left Riemann sum is an overestimate.”
Connected topics
Videos
Check yourself
5 questions on 6.2 Approximating Areas with Riemann Sums. Pick an answer to see if you got it, and why.
Let f(x) = √(1 + x³). Using a midpoint Riemann sum with three subintervals of equal length, what is the approximation of ∫₀⁶ f(x) dx?
| t (seconds) | v(t) (feet per second) |
|---|---|
| 0 | 88 |
| 3 | 84 |
| 4 | 82 |
| 8 | 70 |
| 10 | 60 |
Invented data: velocity of a car as it slows down
A car is moving along a straight road with positive velocity v(t) feet per second, where v is a differentiable function of time t, in seconds. Selected values of v(t) are given in the table. Using a right Riemann sum with the four subintervals indicated by the table, what is the approximation of ∫₀¹⁰ v(t) dt?
On the interval 0 ≤ t ≤ 10, the graph of v is decreasing and concave down. Which of the following approximations of ∫₀¹⁰ v(t) dt, using the subintervals in the table, must be less than the exact value of the integral?
| t (minutes) | R(t) (gallons per minute) |
|---|---|
| 0 | 4 |
| 2 | 7 |
| 5 | 9 |
| 7 | 10 |
| 12 | 11 |
Invented data: rate of water flowing into a tank
Water flows into a tank at a rate of R(t) gallons per minute, where R is a differentiable, strictly increasing function of time t, in minutes. Using a trapezoidal sum with the four subintervals indicated by the table, what is the approximation of ∫₀¹² R(t) dt?
R is strictly increasing, and the graph of R is concave down on the interval 0 ≤ t ≤ 12. Which of the following statements about approximations of ∫₀¹² R(t) dt is true?
0 of 5 answered