Mathematics Advanced • Year 11 • Module 3 • Lesson 14
The Trapezoidal Rule
HSC-style writing on the trapezoidal rule — including an extended response that compares strip counts, justifies over/underestimate using concavity, and discusses the error trend.
1. Short-answer questions
1.1 Use the trapezoidal rule with 4 strips to approximate ∫₀² x² dx, giving your answer to 2 decimal places. 3 marks Band 3
1.2 A function f has the following tabulated values: x = 0, 0.5, 1.0, 1.5, 2.0 and y = 1.00, 1.41, 2.00, 2.83, 4.00. Use the trapezoidal rule to approximate ∫₀² f(x) dx. 3 marks Band 3-4
1.3 Use the trapezoidal rule with 2 strips to estimate ∫₀² (x³ − 3x) dx. Find f″(x) and use it to determine whether your estimate is an under- or overestimate. 4 marks Band 4-5
Stuck on 1.3? Trapezoidal rule overestimates when f is concave up (f″ > 0) and underestimates when f is concave down (f″ < 0).2. Extended response
2.1 Consider the integral I = ∫₀² (1 + x²) dx.
(a) Compute I exactly by direct integration.
(b) Use the trapezoidal rule with n = 2 strips to estimate I. Call this T₂.
(c) Use the trapezoidal rule with n = 4 strips to estimate I. Call this T₄.
(d) Compute the errors |I − T₂| and |I − T₄|, and the ratio (error at n = 2) / (error at n = 4). Comment on whether this ratio matches the trapezoidal-rule prediction that error halves to a quarter when n is doubled. Justify your over/underestimate conclusion by computing f″(x). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — I = [x + x³/3]₀² = 2 + 8/3 = 14/3 ≈ 4.667.
Part (b) — 2 marks
• 1 mark — correctly sets h = 1 and lists y at x = 0, 1, 2 as 1, 2, 5.
• 1 mark — T₂ = (1/2)[1 + 2(2) + 5] = 5.
Part (c) — 2 marks
• 1 mark — correctly sets h = 0.5 and lists y at x = 0, 0.5, 1, 1.5, 2 as 1, 1.25, 2, 3.25, 5.
• 1 mark — T₄ = (0.5/2)[1 + 2(1.25 + 2 + 3.25) + 5] = 0.25(19) = 4.75.
Part (d) — 3 marks
• 1 mark — error₂ = |14/3 − 5| = 1/3 ≈ 0.333; error₄ = |14/3 − 4.75| = 1/12 ≈ 0.083.
• 1 mark — ratio = (1/3) ÷ (1/12) = 4; matches the predicted factor of 4 from h² scaling.
• 1 mark — f″(x) = 2 > 0, so f is concave up on [0, 2]; trapezoidal rule overestimates, consistent with T₂ > I and T₄ > I.
Your response:
Stuck on (d)? Compute each estimate's absolute distance from the exact value 14/3, then divide the error-2 by the error-4.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Trapezoidal estimate of ∫₀² x² dx, n = 4 (3 marks)
Sample response. h = (2 − 0)/4 = 0.5. y-values at x = 0, 0.5, 1, 1.5, 2 are 0, 0.25, 1, 2.25, 4. ≈ (0.5/2)·[0 + 2(0.25 + 1 + 2.25) + 4] = 0.25·[0 + 7 + 4] = 0.25 · 11 = 2.75.
Marking notes. 1 mark — correct h. 1 mark — correct y-values. 1 mark — correct application of the formula with answer to 2 dp. Common error: forgetting the factor of 2 on the interior y-values (gives 0.25·[0 + (0.25 + 1 + 2.25) + 4] = 1.875).
1.2 — Trapezoidal estimate from tabulated y-values (3 marks)
Sample response. n = 4 strips (since 5 values), h = (2 − 0)/4 = 0.5. ≈ (0.5/2)·[1.00 + 2(1.41 + 2.00 + 2.83) + 4.00] = 0.25·[1.00 + 2(6.24) + 4.00] = 0.25·[1 + 12.48 + 4] = 0.25 · 17.48 = 4.37.
Marking notes. 1 mark — recognises n = 4 (not 5) from 5 values. 1 mark — correct interior sum 6.24. 1 mark — correct evaluation 4.37. Common error: using n = 5 (which would force h = 0.4 — does not match the table's 0.5 spacing) and getting a wrong width.
1.3 — ∫₀² (x³ − 3x) dx with n = 2 + concavity check (4 marks)
Sample response. h = 1; x = 0, 1, 2; y = x³ − 3x: 0, 1 − 3 = −2, 8 − 6 = 2. ≈ (1/2)·[0 + 2(−2) + 2] = (1/2)·(−2) = −1. f′(x) = 3x² − 3; f″(x) = 6x. On (0, 2), f″(x) > 0 (concave up). Trapezoidal rule overestimates when concave up, so our estimate −1 is an overestimate of the true (signed) integral. (Check: exact = [x⁴/4 − 3x²/2]₀² = 4 − 6 = −2, and indeed −1 > −2 ✓.)
Marking notes. 1 mark — correct h and y-values. 1 mark — correct trapezoidal estimate −1. 1 mark — computes f″(x) = 6x and notes its sign on (0, 2). 1 mark — correctly identifies overestimate with concavity reason. Common error: stating "underestimate" because the answer is negative — sign of the integral is separate from the under/overestimate property.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). I = ∫₀² (1 + x²) dx = [x + x³/3]₀² = (2 + 8/3) − 0 = 6/3 + 8/3 = 14/3 ≈ 4.667. [1 mark.]
Part (b). n = 2: h = (2 − 0)/2 = 1. x = 0, 1, 2; y = 1 + x²: 1, 2, 5. [1 mark — correct h and y-values.] T₂ = (1/2)·[1 + 2(2) + 5] = (1/2)·[1 + 4 + 5] = (1/2)·10 = 5. [1 mark — correct T₂.]
Part (c). n = 4: h = (2 − 0)/4 = 0.5. x = 0, 0.5, 1, 1.5, 2; y = 1 + x²: 1, 1.25, 2, 3.25, 5. [1 mark.] T₄ = (0.5/2)·[1 + 2(1.25 + 2 + 3.25) + 5] = 0.25·[1 + 2(6.5) + 5] = 0.25·[1 + 13 + 5] = 0.25 · 19 = 4.75. [1 mark.]
Part (d). Errors:
|I − T₂| = |14/3 − 5| = |14/3 − 15/3| = 1/3 ≈ 0.333.
|I − T₄| = |14/3 − 4.75| = |14/3 − 19/4| = |56/12 − 57/12| = 1/12 ≈ 0.083.
[1 mark — both errors correct.]
Ratio = (1/3) ÷ (1/12) = 12/3 = 4. The error has shrunk by exactly the predicted factor of 4 when n is doubled from 2 to 4 (h halved, h² quartered). [1 mark — ratio = 4 and matches h²-scaling prediction.]
Concavity check: f(x) = 1 + x², f′(x) = 2x, f″(x) = 2 > 0 for all x. So f is concave up on [0, 2], and the trapezoidal rule overestimates the area under such curves (trapezoid tops sit above the curve). This is consistent with T₂ = 5 > 14/3 and T₄ = 4.75 > 14/3 — both estimates exceed the exact value. [1 mark — correct concavity argument and consistency check.]
Total: 8/8.
Band descriptors for marker.
Band 3: Computes I and T₂ correctly but cannot compute T₄, OR computes both estimates with arithmetic slips, OR fails to discuss errors. ≈ 3-4 marks.
Band 4: Computes I, T₂, T₄, and errors but does not compute the ratio or merely states "error gets smaller" without linking to h². Concavity argument either missing or not connected to over/underestimate. ≈ 5-6 marks.
Band 5: Completes (a)-(c). Computes errors and ratio = 4 in (d) but omits the f″(x) justification or only mentions it without computing f″. ≈ 6-7 marks.
Band 6: All four parts complete, errors computed exactly as fractions, ratio identified as 4 with explicit h²-scaling argument, and concavity verified via f″(x) = 2 > 0 with the consistency T₂, T₄ > I noted explicitly. 8/8.