Mathematics Advanced • Year 11 • Module 3 • Lesson 14
The Trapezoidal Rule
Apply the trapezoidal rule to river flow surveys, traffic counts, sea-level depth profiles, drug clearance from a tabulated function, and farm-block area estimation.
Problem 1 — River cross-section and flow rate
A hydrologist measures depth (m) across a 12 m wide river at evenly spaced stations:
| Distance (m) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Depth (m) | 0 | 1.4 | 2.5 | 3.2 | 2.8 | 1.6 | 0 |
Set up: What are we solving for?
(i) Use the trapezoidal rule with 6 strips to estimate the cross-sectional area in m². 2 marks
(ii) If the river is flowing at an average speed of 0.6 m/s, estimate the volumetric flow rate (m³/s). 2 marks
(iii) The depth profile is concave down on the wider middle of the river. State whether your trapezoidal estimate is likely an under- or overestimate, and why. 2 marks
Stuck? Revisit lesson § Trap 03 — concave-down regions cause the trapezoidal rule to underestimate.Problem 2 — Cars through an intersection
A traffic camera records the rate of cars per minute crossing an intersection at five-minute intervals over 30 minutes of morning peak:
| Time t (min) | 0 | 5 | 10 | 15 | 20 | 25 | 30 |
|---|---|---|---|---|---|---|---|
| Rate r(t) (cars/min) | 20 | 32 | 45 | 48 | 42 | 28 | 15 |
Set up: What are we solving for?
(i) Total cars through is ∫₀³⁰ r(t) dt. Use the trapezoidal rule with all 6 strips to estimate this total. 2 marks
(ii) Re-estimate using only every 10 minutes (i.e. n = 3 strips: data at 0, 10, 20, 30). How much do the two estimates differ, and why does the coarser estimate matter less for "total cars through"? 3 marks
(iii) Comment in one sentence on which approximation should be more accurate, and link this back to the lesson rule "more strips → more accurate". 2 marks
Problem 3 — Cross-section of a harbour channel
A marine engineer sounds the seabed across a 100 m wide channel at 20 m intervals:
| Across (m) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Depth (m) | 0 | 4 | 9 | 11 | 7 | 0 |
Set up: What are we solving for?
(i) Use the trapezoidal rule with 5 strips to estimate the cross-sectional area of the channel. 2 marks
(ii) Dredging will remove 1 m of depth uniformly across the channel. Estimate the volume of sediment removed (assume the channel is 50 m long). 3 marks
(iii) Trucks haul 12 m³ each. To the nearest truck, how many truckloads are required? 2 marks
Problem 4 — Total drug exposure (AUC)
A patient is given a painkiller. Blood samples taken at hourly intervals give the concentration C(t) in mg/L:
| t (h) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|
| C (mg/L) | 0 | 9.5 | 14.0 | 12.5 | 9.0 | 5.5 | 3.0 |
The "area under the concentration-time curve" (AUC) is ∫₀⁶ C(t) dt; it is used by pharmacologists as a measure of total exposure (in mg·h/L).
Set up: What are we solving for?
(i) Use the trapezoidal rule with 6 strips to estimate AUC over [0, 6]. 2 marks
(ii) A safe dosing protocol requires AUC to be at most 60 mg·h/L over any 6-hour window. State whether this patient is within the limit and by how much. 2 marks
(iii) A nurse takes only 3 samples (t = 0, 3, 6). Use n = 2 strips to estimate AUC, then compare with (i). Explain in one sentence why the coarse estimate undershoots. 3 marks
Stuck on (iii)? With only 3 samples you skip the peak at t = 2 (14.0 mg/L), so the connecting trapezoid sits well below the true curve.Problem 5 — Area of an irregular farm block
A surveyor measures the width of a farm paddock (perpendicular to a straight road) at evenly spaced points along the road:
| Along road (m) | 0 | 30 | 60 | 90 | 120 | 150 |
|---|---|---|---|---|---|---|
| Width (m) | 40 | 55 | 72 | 80 | 65 | 45 |
Set up: What are we solving for?
(i) Estimate the area of the paddock using the trapezoidal rule with n = 5 strips. 2 marks
(ii) Convert to hectares (1 ha = 10 000 m²) and round to 2 dp. 1 mark
(iii) The farmer wants a more accurate value. Suggest two specific things they could do that would, by the trapezoidal-rule theory, increase the accuracy of the estimate. 3 marks
How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — River cross-section
Set up. Treat depth as f(x), apply the trapezoidal rule with h = 2, then multiply by speed to get flow.
(i) h = 2; interior sum = 1.4 + 2.5 + 3.2 + 2.8 + 1.6 = 11.5. Area ≈ (2/2)·[0 + 2(11.5) + 0] = 1 · 23 = 23 m².
(ii) Flow ≈ area × speed = 23 × 0.6 = 13.8 m³/s.
(iii) If the depth profile is concave down in the middle (banks descend, deepest in middle), the trapezoids' straight tops lie below the curve, so the trapezoidal rule underestimates the true area. The real cross-section (and so the real flow) is a little larger than 23 m² / 13.8 m³/s.
Problem 2 — Cars through (rate data)
Set up. Total cars = ∫ r dt. Apply trapezoidal rule with the full 6-strip and the coarse 3-strip data.
(i) h = 5; interior sum = 32 + 45 + 48 + 42 + 28 = 195. Total ≈ (5/2)·[20 + 2(195) + 15] = 2.5·[20 + 390 + 15] = 2.5 · 425 = 1062.5 cars (≈ 1063 cars).
(ii) n = 3, h = 10; y at t = 0, 10, 20, 30: 20, 45, 42, 15. Interior sum = 45 + 42 = 87. Total ≈ (10/2)·[20 + 2(87) + 15] = 5·[20 + 174 + 15] = 5·209 = 1045 cars. Difference ≈ 17.5 cars (≈ 1.6%). For a "rough total over peak", the small relative gap means the coarse estimate is still defensible — but every individual 5-min count is lost.
(iii) The 6-strip estimate is more accurate because more strips means each trapezoid is narrower and hugs the curve more closely; with only 3 strips you skip the peak at t = 15 (48 cars/min), and the connecting trapezoid sits lower than the true curve, undershooting.
Problem 3 — Channel cross-section
Set up. Apply trapezoidal rule to find current area; multiply by channel length to find current volume; subtract a uniform 1 m layer to find sediment.
(i) h = 20; interior sum = 4 + 9 + 11 + 7 = 31. Area ≈ (20/2)·[0 + 2(31) + 0] = 10 · 62 = 620 m².
(ii) Sediment volume removed = (1 m depth × 100 m width × 50 m length) = 5000 m³. (The 1 m layer covers the full surface area; depth across is uniform now even though the underlying profile varies.)
(iii) 5000 ÷ 12 ≈ 416.7 → 417 truckloads.
Problem 4 — Drug AUC
Set up. AUC = ∫ C dt. Apply trapezoidal rule with the full 6-strip and the coarse 2-strip data.
(i) h = 1; interior sum = 9.5 + 14.0 + 12.5 + 9.0 + 5.5 = 50.5. AUC ≈ (1/2)·[0 + 2(50.5) + 3.0] = 0.5·(0 + 101 + 3) = 0.5 · 104 = 52.0 mg·h/L.
(ii) 52 ≤ 60, so the patient is within the safe AUC limit by 8 mg·h/L.
(iii) With samples at t = 0, 3, 6 (n = 2, h = 3): y values 0, 12.5, 3.0. AUC ≈ (3/2)·[0 + 2(12.5) + 3.0] = 1.5·[25 + 3] = 1.5 · 28 = 42.0 mg·h/L. This is 10 mg·h/L less than the 6-strip estimate. The coarse estimate undershoots because it bypasses the peak concentration at t = 2 (14.0 mg/L) entirely — the straight line from (0, 0) to (3, 12.5) sits well below the actual peak.
Problem 5 — Farm paddock area
Set up. Apply the trapezoidal rule with n = 5, then convert to hectares.
(i) h = 30; interior sum = 55 + 72 + 80 + 65 = 272. Area ≈ (30/2)·[40 + 2(272) + 45] = 15·[40 + 544 + 45] = 15 · 629 = 9435 m².
(ii) 9435 ÷ 10 000 = 0.94 ha (2 dp).
(iii) Two specific improvements (any two of):
1. Take more measurements (i.e. use smaller h / more strips) — the trapezoidal error decreases as h² so halving h reduces the error by about a quarter.
2. Take more dense measurements in regions where the boundary changes most (around the maximum width near the middle).
3. Use a more accurate numerical method (e.g. Simpson's rule, which uses quadratic arcs instead of straight tops) once available.
4. Survey using a more precise method (GPS perimeter walk) rather than perpendicular-width transects.