Mathematics Advanced • Year 11 • Module 3 • Lesson 14
The Trapezoidal Rule
Build procedural fluency in approximating definite integrals using the trapezoidal rule — finding h, listing y-values, applying the formula, and assessing over/underestimate.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Complete the trapezoidal rule formula with n strips:
∫ from a to b of f(x) dx ≈ (h/____)[y₀ + ____(y₁ + y₂ + … + y_{n−1}) + y_n], where h = ____________.
Q1.2 True or false (circle one):
(a) With n strips you need n + 1 function values. T / F
(b) Interior y-values are doubled; endpoint y-values are counted once. T / F
(c) The trapezoidal rule overestimates when the curve is concave down. T / F
Q1.3 Approximate ∫₀⁴ x dx using 4 strips (n = 4). State h, the five y-values (y₀ to y₄), and the answer.
h = ____ ; y-values: ____, ____, ____, ____, ____ ; estimate = ____.
2. Worked example — trapezoidal rule with 4 strips for ∫₀² x² dx
Follow each line of algebra. Every step has a reason on the right.
Step 1 — Compute the strip width h.
a = 0, b = 2, n = 4 ⇒ h = (2 − 0)/4 = 0.5
Reason: divide the interval [a, b] into n equal subintervals.
Step 2 — Compute the five y-values y₀, y₁, y₂, y₃, y₄.
x: 0 0.5 1 1.5 2
y: 0 0.25 1 2.25 4
Reason: evaluate f(x) = x² at each x-value.
Step 3 — Apply the formula.
≈ (0.5/2) · [0 + 2(0.25 + 1 + 2.25) + 4]
= 0.25 · [0 + 2(3.5) + 4] = 0.25 · 11 = 2.75
Reason: trapezoidal rule with endpoint coefficients 1 and interior coefficients 2.
Step 4 — Compare with the exact value.
Exact: ∫₀² x² dx = 8/3 ≈ 2.667
Reason: 2.75 > 2.667 — the trapezoidal rule overestimates because y = x² is concave up.
Conclusion. Estimate is 2.75, an overestimate by ≈ 0.083 (curve concave up).
3. Faded example — fill in the missing steps
Use the trapezoidal rule with 4 strips to approximate ∫₁³ 1/x dx. 3 marks
Step 1 — Find h:
h = (____ − ____)/____ = ____
Step 2 — Find the 5 y-values y = 1/x:
x: 1, 1.5, 2, 2.5, 3
y: 1, ____, ____, ____, ____
Step 3 — Apply formula:
≈ (____/2) · [1 + 2(____ + ____ + ____) + ____ ] = ____
4. Graduated practice — trapezoidal approximations
State h, list the y-values, then apply the formula.
Foundation — clean strips (4 questions)
| Q | Integral | Strips n | Estimate |
|---|---|---|---|
| 4.1 1 | ∫₀⁴ x dx | 4 | |
| 4.2 1 | ∫₀² (2x + 1) dx | 4 | |
| 4.3 1 | ∫₀⁴ x² dx | 4 | |
| 4.4 1 | ∫₀² x³ dx | 4 |
Standard — from a table or a formula (6 questions)
Set up h, list the y-values, then apply the formula.
4.5 Use 3 strips to approximate ∫₁⁴ x² dx. 2 marks
4.6 A curve has values (0, 1), (1, 3), (2, 5), (3, 7). Approximate ∫₀³ y dx. 2 marks
4.7 Use the trapezoidal rule with 5 function values to approximate ∫₀² eˣ dx. (Use e⁰ = 1, e^{0.5} ≈ 1.65, e¹ ≈ 2.72, e^{1.5} ≈ 4.48, e² ≈ 7.39.) Round to 3 sig fig. 2 marks
4.8 Use 4 strips to approximate ∫₀¹ √(1 − x²) dx. (Compute y at x = 0, 0.25, 0.5, 0.75, 1.) 2 marks
4.9 The depth d (m) of a river at evenly spaced stations across its width is:
distance (m): 0, 2, 4, 6, 8, 10, 12; depth (m): 0, 1.4, 2.5, 3.2, 2.8, 1.6, 0.
Use the trapezoidal rule to estimate the river's cross-sectional area. 2 marks
4.10 Use 4 strips to approximate ∫₁⁵ 1/x dx. (Compare with the exact value ln 5 ≈ 1.609.) 2 marks
Extension — concavity check + accuracy (2 questions)
4.11 Use the trapezoidal rule with 2 strips to approximate ∫₀² (x³ − 3x) dx. Then state, with reasons, whether your answer is an overestimate or underestimate of the exact value (use f″(x) to decide). 3 marks
4.12 Re-do 4.3 (∫₀⁴ x² dx) using 8 strips. Compute the approximation, and compare the error (against the exact value 64/3) for n = 4 and n = 8. By roughly what factor did the error decrease? 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Trapezoidal rule formula
(h/2)·[y₀ + 2(y₁ + y₂ + … + y_{n−1}) + y_n], with h = (b − a)/n.
Q1.2 — True/false
(a) T. (b) T. (c) F — trapezoidal underestimates when concave down (the straight tops sit below the curve).
Q1.3 — ∫₀⁴ x dx with n = 4
h = 1; y-values: 0, 1, 2, 3, 4. Estimate = (1/2)·[0 + 2(1 + 2 + 3) + 4] = (1/2)·[0 + 12 + 4] = (1/2)·16 = 8. (Exact value is also 8 — trapezoidal rule is exact for linear functions.)
Q3 — Faded example: ∫₁³ 1/x dx with n = 4
h = (3 − 1)/4 = 0.5. y-values: 1, 2/3, 1/2, 2/5, 1/3. ≈ (0.5/2)·[1 + 2(2/3 + 1/2 + 2/5) + 1/3] = 0.25·[1 + 2·(20/30 + 15/30 + 12/30) + 1/3] = 0.25·[1 + 2·(47/30) + 1/3] = 0.25·[1 + 47/15 + 5/15] = 0.25·(15/15 + 47/15 + 5/15) = 0.25·(67/15) = 67/60 ≈ 1.117.
Q4.1 — ∫₀⁴ x dx, n = 4
h = 1; y: 0, 1, 2, 3, 4. ≈ (1/2)·[0 + 2(1 + 2 + 3) + 4] = 8 (exact).
Q4.2 — ∫₀² (2x + 1) dx, n = 4
h = 0.5; y: 1, 2, 3, 4, 5. ≈ (0.5/2)·[1 + 2(2 + 3 + 4) + 5] = 0.25·[1 + 18 + 5] = 0.25 · 24 = 6 (exact).
Q4.3 — ∫₀⁴ x² dx, n = 4
h = 1; y: 0, 1, 4, 9, 16. ≈ (1/2)·[0 + 2(1 + 4 + 9) + 16] = (1/2)·[0 + 28 + 16] = (1/2)·44 = 22. (Exact: 64/3 ≈ 21.33 — overestimate, as x² is concave up.)
Q4.4 — ∫₀² x³ dx, n = 4
h = 0.5; y: 0, 0.125, 1, 3.375, 8. ≈ (0.5/2)·[0 + 2(0.125 + 1 + 3.375) + 8] = 0.25·[0 + 9 + 8] = 0.25 · 17 = 4.25. (Exact: 4 — overestimate.)
Q4.5 — ∫₁⁴ x² dx, n = 3
h = 1; y: 1, 4, 9, 16. ≈ (1/2)·[1 + 2(4 + 9) + 16] = (1/2)·[1 + 26 + 16] = (1/2)·43 = 21.5. (Exact: 63/3 = 21.)
Q4.6 — Tabulated values (0,1), (1,3), (2,5), (3,7)
h = 1; y: 1, 3, 5, 7. ≈ (1/2)·[1 + 2(3 + 5) + 7] = (1/2)·[1 + 16 + 7] = 12. (Exact, since data is linear.)
Q4.7 — ∫₀² eˣ dx, n = 4 (using given values)
h = 0.5; y: 1, 1.65, 2.72, 4.48, 7.39. Sum interior: 1.65 + 2.72 + 4.48 = 8.85. ≈ (0.5/2)·[1 + 2(8.85) + 7.39] = 0.25·[1 + 17.7 + 7.39] = 0.25 · 26.09 = 6.52 (3 sig fig). (Exact: e² − 1 ≈ 6.389; overestimate, since eˣ is concave up.)
Q4.8 — ∫₀¹ √(1 − x²) dx, n = 4
h = 0.25; y: √1 = 1; √(1 − 0.0625) = √0.9375 ≈ 0.968; √(1 − 0.25) = √0.75 ≈ 0.866; √(1 − 0.5625) = √0.4375 ≈ 0.661; √0 = 0. ≈ (0.25/2)·[1 + 2(0.968 + 0.866 + 0.661) + 0] = 0.125·[1 + 2(2.495) + 0] = 0.125·[1 + 4.990] = 0.125 · 5.990 ≈ 0.749. (Exact = π/4 ≈ 0.785; underestimate, as √(1 − x²) is concave down.)
Q4.9 — River cross-section
h = 2 (m); y-values: 0, 1.4, 2.5, 3.2, 2.8, 1.6, 0. ≈ (2/2)·[0 + 2(1.4 + 2.5 + 3.2 + 2.8 + 1.6) + 0] = 1·[0 + 2(11.5) + 0] = 23 m².
Q4.10 — ∫₁⁵ 1/x dx, n = 4
h = 1; y: 1, 0.5, 1/3, 0.25, 0.2. ≈ (1/2)·[1 + 2(0.5 + 1/3 + 0.25) + 0.2] = (1/2)·[1 + 2(1.0833) + 0.2] = (1/2)·[1 + 2.1667 + 0.2] = (1/2)·3.3667 ≈ 1.683. (Exact: ln 5 ≈ 1.609. Overestimate, as 1/x is concave up on (0, ∞).)
Q4.11 — ∫₀² (x³ − 3x) dx with n = 2
h = 1; x: 0, 1, 2; y = x³ − 3x: 0, −2, 2. ≈ (1/2)·[0 + 2(−2) + 2] = (1/2)·(−2) = −1. f″(x) = 6x. On (0, 2): f″ > 0 except at x = 0, so the curve is concave up — trapezoids sit above the curve, so the estimate is an overestimate of the signed integral. (Exact: [x⁴/4 − 3x²/2]₀² = 4 − 6 = −2. Our estimate −1 > −2 ✓.)
Q4.12 — ∫₀⁴ x² dx with n = 8 vs n = 4
n = 8: h = 0.5; x = 0, 0.5, 1, …, 4; y = x²: 0, 0.25, 1, 2.25, 4, 6.25, 9, 12.25, 16. Sum interior: 0.25 + 1 + 2.25 + 4 + 6.25 + 9 + 12.25 = 35. ≈ (0.5/2)·[0 + 2(35) + 16] = 0.25 · (0 + 70 + 16) = 0.25 · 86 = 21.5. Exact 64/3 ≈ 21.333. Error at n = 4: 22 − 21.333 ≈ 0.667. Error at n = 8: 21.5 − 21.333 ≈ 0.167. Error roughly quartered when n doubles (ratio 0.667/0.167 ≈ 4). The trapezoidal rule error scales as h² ∝ 1/n², so doubling n reduces error by ≈ 4.