AP Calculus AB · Lesson 1

Riemann sums and the definite integral

Know Know it

If a rate is constant, amount = rate × time. When the rate changes, split the time into short pieces, treat the rate as constant on each piece, and add. That sum is a Riemann sum.

Left and right sums use the function’s value at the left or right end of each subinterval as the height. A midpoint sum uses the value in the middle. A trapezoidal sum averages the two end values on each piece: f(xk−1)+f(xk)2 Δxk\frac{f(x_{k-1}) + f(x_k)}{2}\,\Delta x_k. The widths do not have to be equal; in a table, use the actual gaps between the listed inputs.

The definite integral is the limit of Riemann sums as the widths shrink: ∫abf(x) dx=lim⁡n→∞∑k=1nf(xk) Δx\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k)\,\Delta x, with Δx=b−an\Delta x = \frac{b - a}{n} and xk=a+k Δxx_k = a + k\,\Delta x. Geometrically, it is the area above the x-axis minus the area below it.

Units: the units of ff times the units of xx. A rate in liters per minute, integrated over minutes, gives liters.

When a sum is too low or too high
SumToo low (underestimate) when f is…Too high (overestimate) when f is…
Leftincreasingdecreasing
Rightdecreasingincreasing
Midpointconcave upconcave down
Trapezoidalconcave downconcave up

Calculator On the exam

On the exam, a graphing calculator is required for one part of the multiple-choice section (13 of the 42 questions, though not all of them need it) and for the first 2 of the 6 free-response questions. In the other parts, no calculator is allowed. You may bring an approved handheld graphing calculator; the exam’s testing app also has a graphing calculator, available only in the calculator parts. Check the current calculator policy before test day.

In a calculator part, a graphing calculator can evaluate a definite integral numerically, so write the integral with its limits and then the value. Without a calculator, use a Riemann sum, geometry (areas of rectangles, triangles and trapezoids) or an antiderivative, and show the setup.

Trap Left and right depend on direction; trapezoids depend on bending

Whether a left or right sum is too high or too low depends on whether f is increasing or decreasing. Whether a trapezoidal or midpoint sum is too high or too low depends on concavity. Do not mix up the two reasons in a justification.

Worked example Easy

Using 3 subintervals of equal width, what is the left Riemann sum for f(x)=x2f(x) = x^2 on [0,3][0, 3]?

  1. 5Answer
  2. 9
  3. 9.5
  4. 14

How to solve it

  1. Δx=3−03=1\Delta x = \frac{3 - 0}{3} = 1, so the subintervals are [0,1][0, 1], [1,2][1, 2] and [2,3][2, 3].
  2. Left endpoints: f(0)=0f(0) = 0, f(1)=1f(1) = 1, f(2)=4f(2) = 4.
  3. Left sum: 1(0)+1(1)+1(4)=51(0) + 1(1) + 1(4) = 5.

Why each choice is right or wrong

  • A. Correct. The widths are 1 and the left endpoints are 0, 1 and 2, so the sum is 1(0+1+4)=51(0 + 1 + 4) = 5.
  • B. Incorrect. 9 is the exact value of ∫03x2 dx\int_0^3 x^2\,dx, not the left Riemann sum, which is lower because x2x^2 is increasing.
  • C. Incorrect. 9.5 is the trapezoidal sum, the average of the left sum 5 and the right sum 14.
  • D. Incorrect. 14 uses the right endpoints 1, 2 and 3: 1+4+91 + 4 + 9.

Explain Explain it

Task verbs: calculate, explain

Show the sum, not just the total. Write each width times each height, such as 2(10)+4(16)+…2(10) + 4(16) + \dots, so a reader can check your choice of endpoints and widths.

Explain an over- or underestimate with the property that decides it: "Because the rate is concave down, each trapezoid’s top edge lies below the graph, so the trapezoidal sum is an underestimate."

Worked example Medium

Water flows into a pond at a rate of W(t)W(t) liters per minute. Selected values are in the table.

Inflow rate W(t)W(t)

tt (minutes)02610
W(t)W(t) (liters per minute)10162215

Using a trapezoidal sum with the three subintervals in the table, what is the estimate of ∫010W(t) dt\int_0^{10} W(t)\,dt?

  1. 50.5 liters
  2. 172 liters
  3. 176 litersAnswer
  4. 180 liters

How to solve it

  1. The widths are 2, 4 and 4 minutes.
  2. Trapezoids: 2⋅13=262 \cdot 13 = 26, 4⋅19=764 \cdot 19 = 76, 4⋅18.5=744 \cdot 18.5 = 74.
  3. Total: 26 + 76 + 74 = 176 liters of water, from t = 0 to t = 10.

Why each choice is right or wrong

  • A. Incorrect. 50.5 averages the end values on each piece but treats every width as 1; the widths are 2, 4 and 4 minutes.
  • B. Incorrect. 172 is the left Riemann sum, 2(10)+4(16)+4(22)2(10) + 4(16) + 4(22), not the trapezoidal sum.
  • C. Correct. 2⋅10+162+4⋅16+222+4⋅22+152=26+76+74=1762\cdot\frac{10 + 16}{2} + 4\cdot\frac{16 + 22}{2} + 4\cdot\frac{22 + 15}{2} = 26 + 76 + 74 = 176 liters.
  • D. Incorrect. 180 is the right Riemann sum, 2(16)+4(22)+4(15)2(16) + 4(22) + 4(15), not the trapezoidal sum.

Explain Your turn

People enter a stadium at a rate of E(t)E(t) people per minute, where tt is minutes after the gates open, and EE is increasing for 0≤t≤300 \leq t \leq 30. A left Riemann sum with three equal subintervals is used to estimate ∫030E(t) dt\int_0^{30} E(t)\,dt. Say what the integral means in this setting, and explain whether the left-sum estimate is too high or too low.

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Compare with a model answer

A full answer does this

  • Says the integral is the total number of people who enter the stadium during the first 30 minutes after the gates open
  • Says the left-sum estimate is too low (an underestimate)
  • Gives the reason: E is increasing, so each left-endpoint value is the smallest value of E on its subinterval

Model answer

∫030E(t) dt\int_0^{30} E(t)\,dt is the total number of people who enter the stadium during the first 30 minutes after the gates open. The left-sum estimate is too low, an underestimate. Because EE is increasing, each rectangle uses the value of EE at the left end of its subinterval, which is the smallest value of EE on that piece, so every rectangle lies under the graph.

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