Trapezoidal, midpoint and Simpson's rules
Problem 4.676 · medium
Approximate \( \displaystyle \int_{1}^{2} \frac{1}{x}\, dx \) using the Simpson's rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{6} = \frac{1}{6} \]Δx = (b − a)/n.✓ Proved
- The Simpson's rule uses the points x = 1, 7/6, 4/3, 3/2, 5/3, 11/6, 2 with weights 1, 4, 2, 4, 2, 4, 1, all times 1/18.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{2} + 1 \cdot 1 + 2 \cdot 3 \cdot \frac{1}{5} + 2 \cdot 3 \cdot \frac{1}{4} + 4 \cdot 6 \cdot \frac{1}{11} + 4 \cdot 2 \cdot \frac{1}{3} + 4 \cdot 6 \cdot \frac{1}{7}}{18} = \frac{14411}{20790} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{1}^{2} \frac{1}{x}\, dx = \ln{\left(2 \right)} \]The exact value, ≈ 0.693147, for comparison: the error is about 2.26e-05.✓ Proved
Answer \( S_{6} = \frac{14411}{20790} \approx 0.693170 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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: fail (error) — The weighted sum in step 3 misapplies Simpson’s weights: the coefficients for f(7/6), f(4/3), f(3/2), f(5/3), f(11/6) are wrong, leading to an incorrect numerical result.qwen3.6:27b-mlx: fail (error) — The function values in step 3 are incorrect; for example, f(7/6) is 6/7, not 1/2. Additionally, the weights listed in step 2 are inconsistent with the standard Simpson's rule formula (which uses Δx/3, not 1/18) and the calculation in step 3.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The function values in step 3 are incorrect; for example, f(7/6) is 6/7, not 1/2. Additionally, the weights listed in step 2 are inconsistent with the standard Simpson's rule formula (which uses Δx/3, not 1/18) and the calculation in step 3.gpt-oss:20b: fail (error) 2026-10-08 — The weighted sum in step 3 misapplies Simpson’s weights: the coefficients for f(7/6), f(4/3), f(3/2), f(5/3), f(11/6) are wrong, leading to an incorrect numerical result.gpt-oss:20b: fail (error) 2026-10-08 — The solution misapplies Simpson’s rule by using incorrect function values. For example, at x=1 the integrand 1/x equals 1, yet the calculation uses 1/2. All weighted terms are therefore wrong, leading to an incorrect approximation.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The function values listed in step 3 are incorrect; for example, f(7/6) is 6/7, not 1/2. Additionally, the weights in step 2 are inconsistent with the standard formula (which uses Δx/3 = 1/18), but the primary error is the evaluation of the function at the grid points.
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 2026-10-08 with SymPy 1.14.0.