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Approximating Areas with Riemann Sums

Use each interval width in rectangular or trapezoidal approximations and justify any error direction.

College Board Topic 6.2 · CED p. 114 (PDF p. 119) · CED effective Fall 2020; project course outline 2026–27.

When an exact antiderivative is unavailable—or you only have samples—approximate the integral with small pieces. On [a,b], partition a = x₀ < x₁ < … < xₙ = b. A left sum uses f(xᵢ₋₁)Δxᵢ, a right sum uses f(xᵢ)Δxᵢ, and a midpoint sum uses f((xᵢ₋₁+xᵢ)/2)Δxᵢ. A trapezoid uses the average of its endpoint heights. Unequal widths are fine; calculate every width separately.

A sampled increasing rate
t (min)013
r(t) (L/min)248

Here left = 2(1)+4(2)=10, right = 4(1)+8(2)=20, and trapezoidal = ½(2+4)(1)+½(4+8)(2)=15 liters. For an increasing function, left underestimates and right overestimates. For a concave-up function, trapezoids overestimate and midpoint rectangles underestimate; concave down reverses that comparison. Sampled data alone do not prove behavior between samples. A calculator can add a sum or evaluate an integral, but it cannot supply a missing assumption.

Let's work through it

  1. Use widths 1 and 2, rather than one repeated width. Reason: The partition 0,1,3 is nonuniform.
  2. Add trapezoids: (2+4)/2 · 1 + (4+8)/2 · 2 = 15 L. Reason: Each trapezoid has average endpoint rate times elapsed time.

Approximate a curved graph using widths and heights

For a continuous function supplied only as a curved graph, a definite integral usually cannot be replaced by an exact triangle or circle area. Choose a partition a = x₀ < x₁ < … < xₙ = b. On interval i, the width is Δxᵢ = xᵢ − xᵢ₋₁. The left, right and midpoint estimates are sums of width times the corresponding graph height. The trapezoidal estimate is Σ Δxᵢ[f(xᵢ₋₁) + f(xᵢ)]/2. Use each interval’s own width on a nonuniform partition.

A graph reading is approximate. Write ∫ₐᵇ f(x) dx ≈ T, rather than an equality, unless an exact model justifies it. Include units: height in liters per minute times width in minutes gives liters. A table can make graph readings accessible without changing their approximate status.

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Separate graph-reading uncertainty from quadrature error

Two errors can enter a graph estimate. Quadrature error comes from replacing the curve by rectangles or chords. Reading error comes from approximating the heights. For a decreasing function, exact left heights overestimate and exact right heights underestimate its integral. For a concave-down function, exact trapezoids underestimate and exact midpoint rectangles overestimate; concave-up functions reverse these last two directions. These statements concern exact heights and the behavior on the entire relevant interval.

If every displayed height has error at most ε and a rule uses nonnegative weights whose sum is b − a, the resulting estimate differs from the same rule with exact heights by at most ε(b − a). This does not bound the rule’s quadrature error. Rounded trapezoid and midpoint sums therefore need not bracket the true integral. Refining a partition can improve an approximation, but the difference between two computed estimates is not automatically a rigorous error bound.

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Read and estimate an increasing curved graph

The supplied curve is continuous, strictly increasing and strictly concave down on [0,4]. The height readings, rounded to the nearest 0.1, are f(0) ≈ 1.0, f(0.5) ≈ 1.7, f(1) ≈ 2.0, f(1.5) ≈ 2.2, f(2) ≈ 2.4, f(2.5) ≈ 2.6, f(3) ≈ 2.7, f(3.5) ≈ 2.9 and f(4) ≈ 3.0. Each reading differs from the actual height by at most 0.05. Estimate ∫₀⁴ f(x) dx using four equal subintervals.

An increasing, concave-down curve on 0 ≤ x ≤ 4 with partitions at each integer and midpoint marks at 0.5, 1.5, 2.5 and 3.5. Approximate heights are supplied in an adjacent labeled table: 1.0, 1.7, 2.0, 2.2, 2.4, 2.6, 2.7, 2.9 and 3.0 at successive half-integers.
A curved graph supplies approximate heights, not exact geometric areas.
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An increasing, concave-down curve on 0 ≤ x ≤ 4 with partitions at each integer and midpoint marks at 0.5, 1.5, 2.5 and 3.5. Approximate heights are supplied in an adjacent labeled table: 1.0, 1.7, 2.0, 2.2, 2.4, 2.6, 2.7, 2.9 and 3.0 at successive half-integers.

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  1. The partition 0,1,2,3,4 has four widths of 1. The left heights give L₄ = 1(1.0 + 2.0 + 2.4 + 2.7) = 8.1. The right heights give R₄ = 1(2.0 + 2.4 + 2.7 + 3.0) = 10.1.
  2. Trapezoids use the average of adjacent endpoint heights: T₄ = (L₄ + R₄)/2 = 9.1. Midpoints give M₄ = 1(1.7 + 2.2 + 2.6 + 2.9) = 9.4.
  3. With exact heights, increasing behavior gives L₄ < ∫₀⁴ f < R₄, while concave-down behavior gives T₄ < ∫₀⁴ f < M₄. The rounded sums are not those exact-height sums.
  4. Each rule has total weight 4, so rounding can change it by at most 4(0.05) = 0.20. The graph supports estimates near 9.1 and 9.4, not an exact integral or a guaranteed interval [9.1,9.4]. Monotonicity with the uncertainty allows the broader bound 7.9 < ∫₀⁴ f < 10.3.

Answer and check

L₄ = 8.1, R₄ = 10.1, T₄ = 9.1 and M₄ = 9.4 from displayed readings. The integral remains approximate; each estimate has up to 0.20 reading uncertainty in addition to quadrature error.

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Transfer the rule to a formula and a verbal rate

A rate is r(t) = 1 + t² liters per minute on 0 ≤ t ≤ 2. Separately, another continuous rate is described as positive and strictly concave up on [0,2], with exact readings 1,2,4 liters per minute at t = 0,1,2.

  1. For the formula, two midpoint rectangles have width 1 and heights r(0.5) = 1.25 and r(1.5) = 3.25, giving M₂ = 4.5 liters. Since r is concave up, this midpoint estimate is an underestimate.
  2. The formula can check that direction: ∫₀² (1 + t²) dt = [t + t³/3]₀² = 14/3 liters, which is greater than 4.5 liters.
  3. For the verbally described rate, the exact endpoint values give T₂ = 1(1 + 2)/2 + 1(2 + 4)/2 = 4.5 liters. The stated whole-interval strict concavity puts each chord above the graph, so the trapezoidal estimate is an overestimate. The three samples alone would not establish that concavity.

Answer and check

Formula: M₂ = 4.5 liters < 14/3 liters. Verbal concave-up rate with the stated exact readings: T₂ = 4.5 liters is an overestimate.

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Question 1 of 20 self-assessed

Question 1

For decreasing f, which endpoint sum underestimates?

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Retrieval 1

For decreasing f, which endpoint sum underestimates?

Worked answer and reason

The right sum.

Right endpoints give lower heights on each subinterval.

Retrieval 2

Can you compute a midpoint sum from endpoint data alone?

Worked answer and reason

Usually no.

You need values at the actual midpoints, unless an additional model is provided.

Catch the tempting mistake

More decimal places fix an inaccurate model.
Numerical precision and approximation error are different. Record the partition and the assumptions before rounding.
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For calculus, set the angle selector to Radians. This scientific calculator evaluates arithmetic and functions. It does not graph arbitrary functions, find numerical roots, evaluate numerical derivatives, or compute definite integrals. Those tasks need an approved graphing tool; the May 2027 exam permits approved handheld graphing calculators and Bluebook’s built-in Desmos on calculator-required parts. A graph or decimal estimate helps you check a conclusion; your setup, hypotheses, units and justification still do the explaining.

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Estimate a supplied curved graph and qualify the result

Calculator optional

The supplied graph of f is continuous, strictly decreasing and strictly concave down throughout [0,4]. The accessible readings, rounded to the nearest 0.1, are: x = 0,0.5,1,1.5,2,2.5,3,3.5,4; f(x) ≈ 4.0,4.0,3.9,3.7,3.5,3.2,2.9,2.5,2.0 in that order. Each displayed height has at most 0.05 reading error. No formula for this curve is supplied.

A decreasing, concave-down curve on 0 ≤ x ≤ 4 with integer partitions and midpoint marks. Rounded heights at x = 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5 and 4 are 4.0, 4.0, 3.9, 3.7, 3.5, 3.2, 2.9, 2.5 and 2.0.
Use the displayed readings for numerical estimates; each height has at most 0.05 reading error.
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A decreasing, concave-down curve on 0 ≤ x ≤ 4 with integer partitions and midpoint marks. Rounded heights at x = 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5 and 4 are 4.0, 4.0, 3.9, 3.7, 3.5, 3.2, 2.9, 2.5 and 2.0.

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(a) Using the supplied curved graph readings, write the partition and width for four equal subintervals of [0,4]. Calculate left, right, midpoint and trapezoidal estimates of ∫₀⁴ f(x) dx, displaying the height-and-width expressions. Explain why these numbers are estimates rather than exact areas.

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(b) Compute the two-subinterval trapezoidal estimate using x = 0,2,4 and compare it with T₄. State the left/right and trapezoid/midpoint error directions for exact heights, using the supplied monotonicity and concavity. Bound the effect of reading uncertainty on each four-subinterval estimate. Decide whether [13.3,13.4] is a guaranteed interval containing the integral, and whether |T₄ − T₂| is a rigorous error bound.

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(c) As an analytic transfer, let r(t) = 1 + t² liters per minute for 0 ≤ t ≤ 2. Use two equal midpoint subintervals to approximate the accumulated volume ∫₀² r(t) dt. Use the behavior of r to determine whether that midpoint estimate is an underestimate or overestimate, and check using an exact integral.

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Diagnose an estimate from a verbal rate description

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A continuous flow rate is positive and strictly concave up on the entire time interval [0,2] minutes. Its exact values at t = 0,1,2 are 1,2,4 liters per minute. Use one trapezoid on each of [0,1] and [1,2].

(a) Which choice gives the two-trapezoid estimate of the accumulated volume and a valid comparison with the true integral? Calculate the estimate and justify the direction from the stated whole-interval behavior.

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For the teacher
  • Compare graph, formula, table, and verbal rate descriptions. Use a nonuniform table deliberately. Have students justify bounds before using the calculator.

Unit planning and formative guidance →

Glossary and deeper reading · 7 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

  • Antiderivative

    An antiderivative of a function is another function whose derivative equals the original function on an interval. Two antiderivatives on the same interval differ by a constant. For example, x² is an antiderivative of 2x, and x²+7 is another. An antiderivative is a function, whereas a definite integral is an accumulated value.

    Read more: 6.8 · Antiderivative →

  • Integral

    Integration can mean accumulating a function over an interval or finding an antiderivative. A definite integral gives a number, while an indefinite integral represents a family of antiderivatives. For example, integrating 2x indefinitely gives x²+C, and integrating it from 0 to 2 gives 4. The Fundamental Theorem of Calculus connects these two operations under its hypotheses.

    Read more: 6.7 · Integral →

  • Function

    A mathematical function assigns exactly one output to each input in its domain. Inputs can share an output, but one input cannot have two different outputs. For example, f(x)=x² assigns the output 9 to both 3 and −3. Function notation names the rule and its input; f(3) means the output when the input is 3.

  • Concavity

    Concavity describes how a graph’s slope changes as the input increases. A graph is concave up where its derivative increases and concave down where its derivative decreases. For example, x² is concave up because its derivative 2x increases. A positive second derivative establishes concave-up behavior on an interval, while a negative one establishes concave-down behavior.

    Read more: 5.6 · Concavity →

  • Continuity

    A function is continuous at an input when its value exists, its finite limit exists, and those two values agree. At a domain endpoint, continuity uses the relevant one-sided approach. For example, x² is continuous at 3 because both its value and its nearby limit are 9. Continuity on an interval requires the appropriate condition at every input in that interval.

    Read more: 1.11 · Continuity →

  • Definite integral

    A definite integral accumulates a function over a specified interval as a limit of Riemann sums when that limit exists. Its sign includes the function’s sign and the orientation of the limits. For example, the integral of the constant −2 from 0 to 3 is −6, although the geometric area is 6. In a rate model, the result has rate units multiplied by input units.

    Read more: 6.3 · Definite integral →

  • Interval

    An interval is a set of real numbers containing every number between any two of its members. Parentheses exclude an endpoint, while square brackets include it. For example, (1,4] includes 4 and every number between 1 and 4, but excludes 1. An infinite endpoint always uses a parenthesis because infinity is not a real endpoint.

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