Lesson 27: Riemann Sums & Approximating Area

(a) Opening Question

A car's speedometer is read every few seconds during a 12-second test. The car never slows down. Here is the data:

You want to estimate the total distance traveled, which is the "area under the velocity curve." You only have these five readings, so you decide to approximate with four rectangles, each 3 seconds wide.

Question: If you build each rectangle using the velocity at the left edge of its 3-second interval, do you get an estimate that is too big or too small compared to the true distance? Compute that estimate.

Take 60 seconds before reading on. Hint: the car is always speeding up. On each 3-second window, the left-edge speed is the slowest speed the car reaches during that window — so each rectangle sits below the real curve.

(b) Core Concepts

The big idea: rectangles approximate area

We want the area under the graph of a function f between x = a and x = b — the region bounded by the curve, the x-axis, and the vertical lines x = a and x = b. Usually we cannot find this area with simple geometry, so we approximate it by slicing the interval [a, b] into thin strips and capping each strip with a rectangle whose area we can compute.

Split [a, b] into n subintervals. If they all have the same width, the partition is uniform and that common width is

Δx = (b - a) / n

A Riemann sum is the total area of the approximating rectangles:

Σⁿᵢ₌₁ f(xᵢ*) · Δx

Here xᵢ is the sample point in the i-th subinterval where we measure the height. The whole game is which* sample point you choose.

Three classic choices of height

Take f(x) = x² on [0, 4] with n = 4, so Δx = 1 and the grid points are x = 0, 1, 2, 3, 4.

Left Riemann sum (LRAM) — height = value at the left endpoint of each subinterval. Sample points: 0, 1, 2, 3.

LRAM = 1·[f(0) + f(1) + f(2) + f(3)] = 1·[0 + 1 + 4 + 9] = 14

Right Riemann sum (RRAM) — height = value at the right endpoint. Sample points: 1, 2, 3, 4.

RRAM = 1·[f(1) + f(2) + f(3) + f(4)] = 1·[1 + 4 + 9 + 16] = 30

Midpoint Riemann sum (MRAM) — height = value at the midpoint of each subinterval. Sample points: 0.5, 1.5, 2.5, 3.5.

MRAM = 1·[f(0.5) + f(1.5) + f(2.5) + f(3.5)]
     = 1·[0.25 + 2.25 + 6.25 + 12.25] = 21

• Smooth upward-curving parabola from (0,0) to (4,16)
• Left panel labeled LRAM: four rectangles each width 1, tops at heights f(0)=0, f(1)=1, f(2)=4, f(3)=9; rectangles sit BELOW the curve (left edges touch curve, gaps above)
• Middle panel labeled RRAM: four rectangles, tops at f(1)=1, f(2)=4, f(3)=9, f(4)=16; rectangles poke ABOVE the curve (right edges touch curve)
• Right panel labeled MRAM: four rectangles, tops at f(0.5)=0.25, f(1.5)=2.25, f(2.5)=6.25, f(3.5)=12.25; midpoint of each top edge touches the curve, small over/under pieces nearly cancel
• Caption: same curve, three sampling choices]

The trapezoidal sum

Instead of flat-topped rectangles, connect consecutive points on the curve with straight segments, forming trapezoids. A trapezoid over a subinterval [xᵢ, xᵢ₊₁] of width Δx has area Δx · (f(xᵢ) + f(xᵢ₊₁)) / 2 — width times the average of the two heights. For a uniform partition, summing telescopes into a tidy formula:

TRAP = (Δx / 2)·[f(x₀) + 2f(x₁) + 2f(x₂) + ··· + 2f(xₙ₋₁) + f(xₙ)]

The endpoints get weight 1; every interior point gets weight 2. For f(x) = x² on [0, 4]:

TRAP = (1/2)·[f(0) + 2f(1) + 2f(2) + 2f(3) + f(4)]
     = (1/2)·[0 + 2 + 8 + 18 + 16] = (1/2)·44 = 22

Notice TRAP is exactly the average of LRAM and RRAM: (14 + 30)/2 = 22. That is always true on a uniform partition, and it is a fast way to check your work.

The true area here is ∫₀⁴ x² dx = 64/3 ≈ 21.33. So LRAM = 14 undershot badly, RRAM = 30 overshot, and MRAM = 21 and TRAP = 22 bracketed the truth tightly.

Overestimate or underestimate?

You can often decide the direction of the error without knowing the exact answer — a favorite AP question. Reason from the shape of f:

• If f is increasing on [a, b]: LRAM underestimates (left edge is the lowest point of each strip) and RRAM overestimates (right edge is the highest).
• If f is decreasing: the reverse — LRAM overestimates, RRAM underestimates.
• If f is concave up (cupping upward): TRAP overestimates (the chord lies above the curve) and MRAM underestimates (the tangent-like top lies below).
• If f is concave down: TRAP underestimates and MRAM overestimates.

Memory hook: a chord of a concave-up curve sits above it, so trapezoids — built from chords — overcount.

The integral as a limit of Riemann sums

As we use more, thinner rectangles, the staircase hugs the curve more tightly. The definite integral is the exact area, defined as the limit of Riemann sums as the number of subintervals grows without bound:

∫ₐᵇ f(x) dx = lim_{n→∞} Σⁿᵢ₌₁ f(xᵢ*) · Δx     (Δx → 0)

When this limit exists (it does for any continuous f), it does not matter whether you sampled left, right, or midpoint — all choices converge to the same number. The approximations of this lesson are the finite steps on the way to that exact value.

Non-uniform partitions (tables)

On the AP exam, data often arrives in a table with subintervals of unequal width. Then there is no single Δx; you must use each interval's own width. The Riemann/trapezoid logic is identical — just compute each rectangle or trapezoid separately and add:

Σ (width of subinterval i) · (chosen height i)

Calculator note (TI-84)

Once you have your sample points, the calculator just does arithmetic — but you can speed up a uniform sum with lists:

TI-84: enter the heights in L1, then  sum(L1) → multiply by Δx
Example RRAM x² [0,4]:  L1 = {1,4,9,16},  sum(L1)=30,  ×Δx=1  →  30

For the exact value to compare against, use fnInt:

TI-84: MATH → 9:fnInt( → fnInt(X², X, 0, 4) → 21.333…

(c) Worked Examples

Example 1 — Riemann sums from a formula (foundation) · [NO CALC]

Problem. For f(x) = x² on [0, 2] with n = 4, find the left and right Riemann sums.

Strategy. Uniform partition: Δx = (2−0)/4 = 0.5. Grid points: 0, 0.5, 1, 1.5, 2.

Solution.

LRAM uses left endpoints 0, 0.5, 1, 1.5:

LRAM = 0.5·[0 + 0.25 + 1 + 2.25] = 0.5·(3.5) = 1.75

RRAM uses right endpoints 0.5, 1, 1.5, 2:

RRAM = 0.5·[0.25 + 1 + 2.25 + 4] = 0.5·(7.5) = 3.75

Justification. Since f(x) = x² is increasing on [0, 2], LRAM = 1.75 is an underestimate and RRAM = 3.75 is an overestimate of ∫₀² x² dx = 8/3 ≈ 2.667, which indeed lies between them.

Example 2 — Table with unequal widths (AP format) · [NO CALC]

Problem. Water flows into a tank. The flow rate R(t), in gallons per minute, is recorded:

Use a left Riemann sum with the four subintervals given by the table to approximate ∫₀¹² R(t) dt.

Strategy. Widths are unequal: 2, 3, 2, 5. A left sum weights each subinterval by R at its left endpoint.

Solution.

∫₀¹² R(t) dt ≈ (2)(10) + (3)(18) + (2)(24) + (5)(28)
            = 20 + 54 + 48 + 140 = 262 gallons

Justification. Watch the widths — the last interval is 5 minutes wide, not 1. A common error is to add R-values and multiply by one Δx; there is no single Δx here.

Example 3 — Trapezoidal sum from the same table · [NO CALC]

Problem. Use a trapezoidal sum with the four subintervals above to approximate ∫₀¹² R(t) dt.

Strategy. Each trapezoid is (width)·(average of the two endpoint heights).

Solution.

∫₀¹² R(t) dt ≈ 2·(10+18)/2 + 3·(18+24)/2 + 2·(24+28)/2 + 5·(28+35)/2
            = 2(14) + 3(21) + 2(26) + 5(31.5)
            = 28 + 63 + 52 + 157.5 = 300.5 gallons

Check. The right sum is (2)(18)+(3)(24)+(2)(28)+(5)(35) = 36+72+56+175 = 339. The trapezoid sum 300.5 is the average of left and right: (262 + 339)/2 = 300.5. ✓

Example 4 — Over/underestimate justification (AP exam level) · [NO CALC]

Problem. A table shows that the values of a function g are g(0)=4, g(2)=7, g(6)=10, g(10)=12. The data suggest g is increasing and concave down. The trapezoidal approximation of ∫₀¹⁰ g(x) dx (using the subinterval widths 2, 4, 4) is 89. Is this an overestimate or underestimate? Justify.

Strategy. Direction of trapezoid error depends on concavity, not on increasing/decreasing.

Solution. 89 is an underestimate.

Justification. Because g is concave down on [0, 10], each chord connecting two points on the graph lies below the curve. The trapezoidal sum is built from those chords, so each trapezoid lies below the region under the curve. Therefore the trapezoidal approximation underestimates ∫₀¹⁰ g(x) dx.

(Trapezoid value check: 2·(4+7)/2 + 4·(7+10)/2 + 4·(10+12)/2 = 11 + 34 + 44 = 89, using the widths 2, 4, 4.)

(d) Common Mistakes

Swapping left and right endpoints. Students grab the first n table values for a right sum or the last n for a left sum. Fix: LRAM drops the last point (x₀…xₙ₋₁); RRAM drops the first point (x₁…xₙ). For an increasing function, the right sum should be the larger number — use that as a sanity check.

Forcing a single Δx on unequal widths. From a table with spacing 2, 3, 2, 5, students compute Δx = 12/4 = 3 and multiply everything by 3. Fix: There is no common Δx when widths differ. Multiply each height by its own subinterval width.

Wrong over/underestimate direction. Students tie the trapezoid's direction to increasing/decreasing. Fix: Endpoint sums (left/right) depend on increasing vs. decreasing; trapezoid and midpoint depend on concavity. Concave up → TRAP over, MRAM under. Concave down → TRAP under, MRAM over.

Forgetting the width entirely. Students add the heights and report that as the area — leaving off Δx (or the widths). Fix: A Riemann sum is always Σ height × width. Units check: gal/min × min = gallons.

Miscounting subintervals. With grid points 0,1,2,3,4 there are 4 subintervals, not 5. Fix: n = number of gaps = (number of points) − 1.

(e) Practice Problems

Short answer. For the table in Problem 7, is the left Riemann sum an overestimate or underestimate of ∫₀⁴⁰ v(t) dt? Justify your answer.

Short answer. A table gives h(1)=4, h(3)=7, h(6)=10, h(10)=12. Compute the trapezoidal approximation of ∫₁¹⁰ h(x) dx using the three subintervals.

Short answer. Write the right Riemann sum for f(x)=2x+1 on [1, 5] with n = 4 in sigma notation, then evaluate it.

Justify. A continuous function g is decreasing and concave up on [0, 6]. Order from smallest to largest: the left sum L, the right sum R, the trapezoid T, and the exact integral I (all with the same uniform partition). Justify the placement of T relative to I.

(f) AP Exam Focus — Free Response

This is a calculator-permitted free-response question (Section II, Part A style).

A tank is being filled with water. The rate at which water enters the tank is given by a differentiable function W(t), measured in gallons per minute, where t is measured in minutes. Selected values of W(t) are given in the table below. The function W is known to be increasing for 0 ≤ t ≤ 20.

| t (minutes) | 0 | 4 | 9 | 15 | 20 |

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

| W(t) (gal/min) | 25 | 32 | 40 | 48 | 52 |

(a) (2 points) Use a left Riemann sum with the four subintervals indicated by the table to approximate ∫₀²⁰ W(t) dt. Show the computations that lead to your answer.

(b) (2 points) Use a trapezoidal sum with the four subintervals indicated by the table to approximate ∫₀²⁰ W(t) dt. Show the computations that lead to your answer.

(c) (2 points) Is the left Riemann sum found in part (a) an overestimate or an underestimate of ∫₀²⁰ W(t) dt? Give a reason for your answer.

(d) (2 points) Using correct units, interpret the meaning of ∫₀²⁰ W(t) dt in the context of this problem.

Total: 8 points

Model Solution

(a) The four subintervals have widths 4, 5, 6, 5. A left sum uses W at the left endpoint of each:

∫₀²⁰ W(t) dt ≈ (4)(25) + (5)(32) + (6)(40) + (5)(48)
            = 100 + 160 + 240 + 240
            = 740 gallons

(b) Each trapezoid contributes (width)·(average of the two endpoint heights):

∫₀²⁰ W(t) dt ≈ 4·(25+32)/2 + 5·(32+40)/2 + 6·(40+48)/2 + 5·(48+52)/2
            = 4(28.5) + 5(36) + 6(44) + 5(50)
            = 114 + 180 + 264 + 250
            = 808 gallons

(c) The left Riemann sum is an underestimate. Because W is increasing on [0, 20], the value of W at the left endpoint of each subinterval is the smallest value W attains on that subinterval. Therefore each rectangle lies below the graph of W, and the sum of their areas is less than ∫₀²⁰ W(t) dt.

(d) ∫₀²⁰ W(t) dt is the total number of gallons of water that enter the tank during the time interval from t = 0 to t = 20 minutes. (Units: gal/min × min = gallons.)

Scoring Commentary

Where students lose points:

• Part (a): Treating the widths as all equal (e.g., multiplying by a single Δx). The widths 4, 5, 6, 5 differ — using 5 throughout is the most common error and earns the setup point only if the per-interval structure is otherwise correct.
• Part (c): Saying "underestimate" with no reason, or giving the reason "because it's a left sum." The reader needs the chain increasing → left endpoint is the lowest value → rectangles below the curve. Tying the answer to concavity here earns nothing.
• Part (d): Omitting units, or describing a rate ("how fast water enters") instead of an accumulated amount. The integral of a rate is a total. "The amount of water in the tank" alone is risky — it is the water that entered over the interval, not the absolute amount in the tank unless a starting amount is given.

2·(4+7)/2 + 3·(7+10)/2 + 4·(10+12)/2 = 11 + 25.5 + 44 = 80.5

Σ⁴ᵢ₌₁ f(1 + i)·(1) = Σ⁴ᵢ₌₁ (2(1+i)+1)·1
= f(2)+f(3)+f(4)+f(5) = 5+7+9+11 = 32