Lesson 32: Approximating Definite Integrals

(a) Opening Question

A sensor records the rate at which water flows into a reservoir, in gallons per hour, at a few moments in time:

Estimate the total amount of water that entered the reservoir between t = 0 and t = 5 hours. You are not given a formula for R(t) — only these three readings.

Take a minute before reading on.

The total water is ∫₀⁵ R(t) dt — the accumulation of the rate over the time interval. But with no formula, you cannot use the Fundamental Theorem of Calculus; there is no antiderivative to evaluate. All you have is discrete data. So you do the only sensible thing: connect the dots and find the area underneath. If you join (0, 30), (2, 38), (5, 56) with straight segments, you get two trapezoids, and their combined area estimates the integral. Notice the widths are not equal — 2 hours, then 3 hours. That single fact is at the heart of nearly every AP table problem, and it is what this lesson is built around.

(b) Core Concepts

Why we approximate

Sometimes a definite integral cannot be evaluated exactly. Either the function has no elementary antiderivative, or — far more common on the AP exam — you are never given a formula at all, only a table of measured values or a graph. In those situations you estimate ∫ₐᵇ f(x) dx by adding up the areas of simple shapes (rectangles or trapezoids) built on the data you do have. This is the same Riemann-sum idea from Lesson 27, but now the subinterval widths are dictated by the table and are usually unequal.

A Riemann sum approximates ∫ₐᵇ f(x) dx by partitioning [a, b] into subintervals and summing (height) × (width) over each one. The methods differ only in which height they use on each subinterval.

Left and right Riemann sums from a table

Partition [a, b] at the table's x-values x₀ < x₁ < ⋯ < xₙ. On the subinterval [xᵢ, xᵢ₊₁] the width is Δxᵢ = xᵢ₊₁ − xᵢ.

• Left Riemann sum: use the function value at the left endpoint of each subinterval as the rectangle height.

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L = Σ f(xᵢ)·Δxᵢ = f(x₀)Δx₀ + f(x₁)Δx₁ + ⋯ + f(xₙ₋₁)Δxₙ₋₁

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• Right Riemann sum: use the value at the right endpoint of each subinterval.

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R = Σ f(xᵢ₊₁)·Δxᵢ = f(x₁)Δx₀ + f(x₂)Δx₁ + ⋯ + f(xₙ)Δxₙ₋₁

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Crucial bookkeeping with unequal widths: each height is multiplied by that subinterval's own width, not by a single common Δx. Pairing the wrong width with a height is the most common table error there is.

The midpoint sum (when the table gives midpoints)

A midpoint Riemann sum uses the function value at the midpoint of each subinterval as the height. You can only use it when the table actually supplies a value at each midpoint. If the table gives values only at the partition endpoints, the midpoint sum is not available — don't invent values by averaging.

The trapezoidal rule

Instead of a flat rectangle on each subinterval, the trapezoidal rule connects consecutive data points with a straight segment, forming a trapezoid whose parallel sides are the two function values. The area of one trapezoid on [xᵢ, xᵢ₊₁] is

Area = ½ · (Δxᵢ) · [f(xᵢ) + f(xᵢ₊₁)]

— one-half the width times the sum of the two heights. The trapezoidal approximation is the total of all these trapezoids:

T = Σ ½·(Δxᵢ)·[f(xᵢ) + f(xᵢ₊₁)]

There is one beautiful shortcut worth knowing: on each subinterval the trapezoid area equals the average of the left and right rectangles, so T = ½(L + R). But on the AP exam, build the trapezoids directly — it is faster and less error-prone with unequal widths.

The trapezoidal rule does NOT require equal widths. Because you compute each trapezoid separately and add, unequal spacing is handled automatically. This is exactly why AP tables can space their times however they like.

Worked trapezoidal approximation (unequal widths). Suppose a function f is recorded as:

The three subintervals have widths Δx = 2, 3, 5. Build a trapezoid on each:

[0, 2]:  ½(2)(3 + 5)  = ½(2)(8)  = 8
[2, 5]:  ½(3)(5 + 8)  = ½(3)(13) = 19.5
[5, 10]: ½(5)(8 + 10) = ½(5)(18) = 45

Add them: T = 8 + 19.5 + 45 = 72.5. So ∫₀¹⁰ f(x) dx ≈ 72.5. Notice each height-pair was multiplied by its own width — using a single Δx here would be wrong.

Comparing left, right, and trapezoid on the same data

Run all three methods on the table above:

Left  L = 3(2) + 5(3) + 8(5)  = 6  + 15 + 40 = 61
Right R = 5(2) + 8(3) + 10(5) = 10 + 24 + 50 = 84
Trap  T = 72.5     (and indeed ½(61 + 84) = 72.5 ✓)

This data is increasing, and the ordering L = 61 < T = 72.5 < R = 84 is no accident. It reflects a general rule.

Over- or underestimate: the two rules to know

For left/right sums — look at increasing/decreasing:

• If f is increasing on the interval, then on each subinterval the left endpoint is the lowest point and the right endpoint is the highest. So the left sum underestimates and the right sum overestimates.
• If f is decreasing, it is reversed: the left sum overestimates and the right sum underestimates.

For the trapezoidal rule — look at concavity:

• If f is concave up on the interval, each straight chord lies above the curve, so each trapezoid contains extra area. The trapezoidal rule overestimates.
• If f is concave down, each chord lies below the curve, so the trapezoidal rule underestimates.

How do you judge concavity from a table? Look at the secant slopes between consecutive points: [f(xᵢ₊₁) − f(xᵢ)]/Δxᵢ. If these slopes are increasing as you move right, the data is concave up; if decreasing, concave down. (This is the discrete shadow of "f'' controls concavity.")

Interpreting the integral and its units

When f(x) is a rate of change, the definite integral ∫ₐᵇ f(x) dx is the net accumulated amount over [a, b]. Its units are the units of f multiplied by the units of x:

[rate] × [time] = [amount]

If R(t) is in gallons per hour and t is in hours, then ∫₀⁵ R(t) dt is in (gal/hr)·(hr) = gallons — the total gallons that flowed in over the 5 hours. Stating the units and a context sentence is exactly what earns the interpretation point on an AP rubric.

Calculator note

On a [CALC] problem these approximations are still done by hand from the table — the calculator only does the arithmetic. There is no built-in "trapezoidal rule from a table" button. (fnInt( integrates a formula exactly; it cannot read your table.) Use the calculator to add ½(2)(3+5) + ½(3)(5+8) + ½(5)(8+10), nothing more.

(c) Worked Examples

Example 1 — Trapezoidal approximation from an unequal-width table (CALC)

Problem. Water flows into a tank at rate R(t), in gallons per hour, where t is in hours. Selected values:

Use a trapezoidal sum with the four subintervals indicated by the table to approximate ∫₀¹⁰ R(t) dt.

Strategy. Four data gaps means four trapezoids, with widths 1, 2, 3, 4. Apply ½(Δt)(R_left + R_right) to each and add.

Solution.

[0, 1]:  ½(1)(20 + 24) = ½(1)(44) = 22
[1, 3]:  ½(2)(24 + 30) = ½(2)(54) = 54
[3, 6]:  ½(3)(30 + 33) = ½(3)(63) = 94.5
[6, 10]: ½(4)(33 + 36) = ½(4)(69) = 138

∫₀¹⁰ R(t) dt ≈ 22 + 54 + 94.5 + 138 = 308.5 gallons

Justification of representation. Each height-pair is multiplied by its own width (1, 2, 3, 4); there is no single Δt. The result, 308.5, carries units (gal/hr)(hr) = gallons — the approximate total volume of water that entered the tank over the 10 hours.

Example 2 — Left vs. right sum + over/underestimate justification (NO CALC)

Problem. A car decelerates. Its speed v(t), in feet per second, is recorded with t in seconds:

(i) Find the left and right Riemann sums for ∫₀¹² v(t) dt. (ii) Which is an overestimate of the true integral? Justify.

Strategy. Widths are 3, 4, 5. The left sum uses values at t = 0, 3, 7; the right sum uses values at t = 3, 7, 12. Then use the increasing/decreasing rule.

Solution.

Left:  60(3) + 52(4) + 40(5) = 180 + 208 + 200 = 588
Right: 52(3) + 40(4) + 25(5) = 156 + 160 + 125 = 441

(ii) The table values decrease as t increases, so v is decreasing on [0, 12]. For a decreasing function the left endpoint is the highest point on each subinterval, so every left rectangle pokes above the curve. Therefore the left sum, 588, is an overestimate. (Correspondingly the right sum, 441, is an underestimate, and the true integral lies between them.)

Example 3 — Interpret the integral with units (NO CALC)

Problem. Using the data of Example 2, ∫₀¹² v(t) dt ≈ 514.5 (the trapezoidal estimate). Using correct units, explain what this integral represents in the context of the car.

Strategy. v(t) is a rate (distance per time), so its integral accumulates distance. Identify units and write a context sentence.

Solution. The units are (ft/sec)·(sec) = feet. Since v(t) is the car's speed, ∫₀¹² v(t) dt is the total distance, in feet, that the car travels from t = 0 to t = 12 seconds. The value ≈ 514.5 means the car travels approximately 514.5 feet over those 12 seconds. (Speed is non-negative here, so distance traveled and displacement coincide.)

Example 4 — Compare which approximation is larger (NO CALC)

Problem. For the increasing data below, a student computes the left sum, the right sum, and the trapezoidal sum for ∫₀¹⁰ R(t) dt. Without re-deriving every number, order the three approximations from smallest to largest, and explain.

Strategy. Use the structural rules, not raw arithmetic. The data increases, and the trapezoid is the average of left and right.

Solution. The values increase, so R is increasing on [0, 10]. Hence the left sum underestimates and the right sum overestimates the true integral. Because the trapezoidal sum is the average of the two, T = ½(L + R), it lies strictly between them. Therefore:

Left sum  <  Trapezoidal sum  <  Right sum

Justification. For an increasing function the left endpoints give the smallest heights and the right endpoints give the largest, so L < R; the trapezoid, being their average, falls in the middle. (Numerically, L = 290, T = 308.5, R = 327 — consistent with the ordering, though the reasoning alone is enough to answer.)

(d) Common Mistakes

Using a single Δx when the widths are unequal. Students see a table and reflexively factor out one common width — e.g., writing T = ½(2)[f(x₀) + 2f(x₁) + 2f(x₂) + f(x₃)] for data spaced 2, 3, 5. That compact formula is only valid for equal widths. Fix: with unequal spacing, build each trapezoid separately, ½(Δxᵢ)(f_left + f_right), using that subinterval's own width, and add.

Getting the over/underestimate direction backward. Two different rules govern direction, and students cross them up. Fix: keep them separate. Left/right depends on increasing vs. decreasing; trapezoid depends on concave up vs. concave down. A trapezoid on a concave-up curve overestimates (the chord sits above the curve) — picture the chord lying above a smiling curve.

Confusing the approximation with the exact value. A trapezoidal sum is an estimate of ∫ₐᵇ f(x) dx, not the integral itself. Writing ∫ₐᵇ f(x) dx = 308.5 (with an equals sign) is wrong; it should be ∫ₐᵇ f(x) dx ≈ 308.5. Fix: use ≈ for every table/graph approximation, and never claim the exact value of an integral you only estimated.

Forgetting the ½ in the trapezoid. Dropping the one-half turns each trapezoid into (Δx)(f_left + f_right), doubling every term and the whole answer. Fix: the trapezoid area is ½ · base · (sum of the two parallel sides); the ½ is not optional. (Sanity check: a trapezoid should be between the left and right rectangles, never larger than both.)

Pairing a height with the wrong width. When the spacing is irregular, students multiply, say, the value at t = 3 by the width of the next gap instead of its own. Fix: write the subintervals out explicitly — [0,1], [1,3], [3,6], [6,10] — and read each width directly off the bracket before multiplying.

(e) Practice Problems

Use this table for problems 1–4 (function f recorded at equally spaced x):

| x | 0 | 2 | 4 | 6 |

|---|---|---|---|---|

| f(x) | 1 | 4 | 9 | 16 |

4 (justification). [NO CALC] The data above is increasing and concave up (the secant slopes 1.5, 2.5, 3.5 increase). State, with reasons, whether the left sum and the trapezoidal sum each over- or underestimate ∫₀⁶ f(x) dx.

12 (short answer). [NO CALC] For the table below, approximate ∫₀⁷ f(x) dx with a trapezoidal sum over the three subintervals. Show each trapezoid.

13 (short answer, justification). [NO CALC] Using the same table as #12, find the left and right Riemann sums. The data is increasing — state which of your two sums is the overestimate and justify why.

14 (short answer, justification). [NO CALC] A student approximates ∫₀⁴ f(x) dx for a function known to be concave up, getting a trapezoidal sum of 30. They write ∫₀⁴ f(x) dx = 30. Identify two things wrong with that statement and correct them.

15 (short answer, interpretation). [NO CALC] At a gas station, fuel is pumped at rate P(t) gallons per minute, t in minutes. A trapezoidal estimate gives ∫₀⁵ P(t) dt ≈ 9.4. State the units of this value and write one sentence interpreting it in context.

(f) AP Exam Focus

This is a calculator-permitted free-response question (AP Section II, Part A style) — the classic rate table FRQ. Budget about 15 minutes. Show the setup for every approximation; the calculator is only for arithmetic.

FRQ. Water is pumped into a reservoir. The rate at which water enters is modeled by a differentiable function G(t), measured in gallons per hour, where t is measured in hours, 0 ≤ t ≤ 8. Selected values of G(t) are given in the table below.

| t (hours) | 0 | 2 | 5 | 8 |

|---|---|---|---|---|

| G(t) (gal/hr) | 30 | 38 | 56 | 86 |

(a) Using a trapezoidal sum with the three subintervals indicated by the table, approximate ∫₀⁸ G(t) dt. Show the computation. (3 points)

(b) Using correct units, interpret the meaning of ∫₀⁸ G(t) dt in the context of this problem. (2 points)

(c) The secant slopes of G between consecutive data points are 4, 6, and 10 gal/hr per hour. Based on this, is your trapezoidal approximation in part (a) an overestimate or an underestimate of ∫₀⁸ G(t) dt? Give a reason for your answer. (2 points)

(d) The total amount of water in the reservoir at time t is W(t), and W(0) = 500 gallons. Use your answer from part (a) to approximate W(8). Explain the relationship you used. (2 points)

Total: 9 points

Model Solution

(a) Three subintervals, widths Δt = 2, 3, 3:

[0, 2]: ½(2)(30 + 38) = ½(2)(68)  = 68
[2, 5]: ½(3)(38 + 56) = ½(3)(94)  = 141
[5, 8]: ½(3)(56 + 86) = ½(3)(142) = 213

∫₀⁸ G(t) dt ≈ 68 + 141 + 213 = 422  gallons

(b) The units are (gal/hr)·(hr) = gallons. ∫₀⁸ G(t) dt represents the total amount of water, in gallons, that is pumped into the reservoir from time t = 0 to t = 8 hours. Its approximate value is 422 gallons.

(c) The secant slopes 4, 6, 10 are increasing, which indicates that G is concave up on [0, 8]. For a concave-up function, each chord used by the trapezoidal rule lies above the graph of G, so every trapezoid overstates the area beneath the curve. Therefore the trapezoidal approximation is an overestimate of ∫₀⁸ G(t) dt.

(d) Since G(t) = W'(t) is the rate at which water enters, by the net-change interpretation ∫₀⁸ G(t) dt = W(8) − W(0) — the integral of the rate gives the total change in the amount. Therefore

W(8) = W(0) + ∫₀⁸ G(t) dt ≈ 500 + 422 = 922  gallons.

So the reservoir holds approximately 922 gallons at t = 8 hours.

Scoring Commentary — where students lose points

• (a) 3 points: 1 for a correct trapezoidal setup using the three given subintervals, 1 for using the correct unequal widths 2, 3, 3, 1 for the answer 422. Trap: collapsing to a single trapezoid ½(8)(30 + 86) = 464, or factoring out one Δt — both lose the width point. Dropping the ½ (giving 844) loses points too.
• (b) 2 points: 1 for correct units (gallons), 1 for a context interpretation as total water pumped in over [0, 8]. Trap: describing it as a rate (gal/hr) or as an average — the integral of a rate is an accumulated amount, not a rate.
• (c) 2 points: 1 for the correct direction (overestimate), 1 for the reason tied to concavity (increasing secant slopes ⇒ concave up ⇒ chords above the curve). Trap: justifying with increasing/decreasing — that governs left/right sums, not the trapezoid. A right answer with the wrong reason earns only the direction point, and a bare "overestimate" with no reason earns nothing for the justification.
• (d) 2 points: 1 for the relationship W(8) = W(0) + ∫₀⁸ G(t) dt, 1 for ≈ 922. Trap: reporting 422 (forgetting to add the initial 500), or trying to "find an antiderivative of G" — there is no formula, so the accumulation must come from part (a).

[0, 2]: ½(2)(12 + 20) = ½(2)(32) = 32
[2, 3]: ½(1)(20 + 26) = ½(1)(46) = 23
[3, 7]: ½(4)(26 + 40) = ½(4)(66) = 132