Trapezoidal, midpoint and Simpson's rules
Problem 4.387 · medium
Approximate \( \displaystyle \int_{1}^{3} \frac{1}{x}\, dx \) using the Simpson's rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The Simpson's rule uses the points x = 1, 4/3, 5/3, 2, 7/3, 8/3, 3 with weights 1, 4, 2, 4, 2, 4, 1, all times 1/9.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{3} + 2 \cdot 3 \cdot \frac{1}{7} + 1 \cdot 1 + 2 \cdot 3 \cdot \frac{1}{5} + 4 \cdot 3 \cdot \frac{1}{8} + 4 \cdot 1 \cdot \frac{1}{2} + 4 \cdot 3 \cdot \frac{1}{4}}{9} = \frac{2077}{1890} \]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 3.30e-04.✓ Proved
Answer \( S_{6} = \frac{2077}{1890} \approx 1.098942 \)
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 the Simpson weights: the second point should have weight 4, the last point weight 1, and the order of terms is incorrect. This produces an incorrect numerical value.qwen3.6:27b-mlx: fail (error) — The solution incorrectly applies Simpson's Rule weights. It uses weights 1, 4, 2, 4, 2, 4, 1 but multiplies the sum by 1/9 instead of the correct factor Δx/3 = 1/9. Wait, Δx = 2/6 = 1/3, so Δx/3 = 1/9. The factor is correct. However, the function values listed in step 3 are incorrect. For x=4/3, f(x)=3/4, not 3/7. For x=5/3, f(x)=3/5, not 3/7. The term '2*(3/7)' appears twice, suggesting the evaluator used x=7/3 and x=5/3 incorrectly or just hallucinated values. Specifically, f(4/3) = 3/4, f(5/3) = 3/5, f(7/3) = 3/7. The solution lists 3/7 where 3/4 should be. This is a calculation error in the setup of the sum.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly applies Simpson's Rule weights. It uses weights 1, 4, 2, 4, 2, 4, 1 but multiplies the sum by 1/9 instead of the correct factor Δx/3 = 1/9. Wait, Δx = 2/6 = 1/3, so Δx/3 = 1/9. The factor is correct. However, the function values listed in step 3 are incorrect. For x=4/3, f(x)=3/4, not 3/7. For x=5/3, f(x)=3/5, not 3/7. The term '2*(3/7)' appears twice, suggesting the evaluator used x=7/3 and x=5/3 incorrectly or just hallucinated values. Specifically, f(4/3) = 3/4, f(5/3) = 3/5, f(7/3) = 3/7. The solution lists 3/7 where 3/4 should be. This is a calculation error in the setup of the sum.gpt-oss:20b: fail (error) 2026-10-05 — The weighted sum in step 3 misapplies the Simpson weights: the second point should have weight 4, the last point weight 1, and the order of terms is incorrect. This produces an incorrect numerical value.qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly calculates the step size as 1/3 instead of 1/3 (which is correct for n=6, wait, (3-1)/6 = 1/3, so delta x is correct). However, the weights in line 2 are listed as 1, 4, 2, 4, 2, 4, 1, which is correct for Simpson's rule. But line 3 computes the sum using incorrect function values and weights. Specifically, it lists terms like 1*(1/3) which implies f(1)=1/3? No, f(1)=1. The term 1*(1/3) suggests the weight 1 is multiplied by 1/3? Or is it f(x_i)? f(1)=1. The term 4*(3/7) suggests f(4/3) = 3/4? No, 3/7 is not 3/4. The values used in the sum (1/3, 3/7, 1, 3/5, 3/8, 1/2, 3/4) do not correspond to 1/x at the specified points (1, 4/3, 5/3, 2, 7/3, 8/3, 3). For example, f(4/3) = 3/4, but the solution uses 3/7. The calculation is fundamentally wrong.gpt-oss:20b: fail (error) 2026-10-05 — The weighted sum in step 3 is incorrect: the function values 1/x at the sample points are mis‑listed (e.g., 1*(1/3) instead of 1*1, and the ordering of the terms does not match the Simpson weights).
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 2026-10-05 with SymPy 1.14.0.