Trapezoidal, midpoint and Simpson's rules
Problem 4.829 · medium
Approximate \( \displaystyle \int_{1}^{3} \frac{1}{x}\, dx \) using the trapezoidal rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 1, 4/3, 5/3, 2, 7/3, 8/3, 3 with weights 1, 2, 2, 2, 2, 2, 1, all times 1/6.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{3} + 2 \cdot 3 \cdot \frac{1}{8} + 2 \cdot 3 \cdot \frac{1}{7} + 1 \cdot 1 + 2 \cdot 1 \cdot \frac{1}{2} + 2 \cdot 3 \cdot \frac{1}{5} + 2 \cdot 3 \cdot \frac{1}{4}}{6} = \frac{2789}{2520} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{1}^{3} \frac{1}{x}\, dx = \ln{\left(3 \right)} \]The exact value, ≈ 1.098612, for comparison: the error is about 8.13e-03.✓ Proved
Answer \( T_{6} = \frac{2789}{2520} \approx 1.106746 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the rule implemented separately in floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly applies the trapezoidal rule formula. The standard formula is T_n = (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)]. The solution uses a divisor of 6 (which is n) instead of 3 (which is Δx/2 = 1/6). This results in an answer that is half the correct value.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution incorrectly applies the trapezoidal rule formula. The standard formula is T_n = (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)]. The solution uses a divisor of 6 (which is n) instead of 3 (which is Δx/2 = 1/6). This results in an answer that is half the correct value.qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states that the trapezoidal rule weights are multiplied by 1/6. The standard formula is (Δx/2) * [f(x_0) + 2f(x_1) + ... +gpt-oss:20b: fail (error) 2026-10-10 — The weighted sum in step 3 mis‑evaluates f(1). It uses 1/3 instead of f(1)=1, so the computed trapezoidal approximation is incorrect.
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/numerical_integration, checked 2026-10-10 with SymPy 1.14.0.