Trapezoidal, midpoint and Simpson's rules
Problem 4.522 · medium
Approximate \( \displaystyle \int_{0}^{1} \frac{1}{x^{2} + 1}\, dx \) using the Simpson's rule with \( \displaystyle n = 4 \).
- \[ 1 \cdot \frac{1}{4} = \frac{1}{4} \]Δx = (b − a)/n.✓ Proved
- The Simpson's rule uses the points x = 0, 1/4, 1/2, 3/4, 1 with weights 1, 4, 2, 4, 1, all times 1/12.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{2} + 1 \cdot 1 + 2 \cdot 4 \cdot \frac{1}{5} + 4 \cdot 16 \cdot \frac{1}{25} + 4 \cdot 16 \cdot \frac{1}{17}}{12} = \frac{8011}{10200} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{0}^{1} \frac{1}{x^{2} + 1}\, dx = \frac{\pi}{4} \]The exact value, ≈ 0.785398, for comparison: the error is about 6.01e-06.✓ Proved
Answer \( S_{4} = \frac{8011}{10200} \approx 0.785392 \)
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 misidentifies the function values and their corresponding Simpson weights. It uses 1*(1/2) instead of 1*f(0)=1, and the order of the terms does not match the weights 1,4,2,4,1. Consequently the reported Simpson approximation is not correct.qwen3.6:27b-mlx: fail (error) — The function values in the weighted sum are incorrect. For example, f(1/2) = 4/5, not 1/2, and f(3/4) = 16/25, not 16/17. The solution uses wrong y-values but somehow arrives at the correct final fraction, likely due to a coincidence or hidden error correction, making the intermediate steps factually wrong.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The weighted sum in step 3 misidentifies the function values and their corresponding Simpson weights. It uses 1*(1/2) instead of 1*f(0)=1, and the order of the terms does not match the weights 1,4,2,4,1. Consequently the reported Simpson approximation is not correct.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The function values in the weighted sum are incorrect. For example, f(1/2) = 4/5, not 1/2, and f(3/4) = 16/25, not 16/17. The solution uses wrong y-values but somehow arrives at the correct final fraction, likely due to a coincidence or hidden error correction, making the intermediate steps factually wrong.qwen3.6:27b-mlx: fail (error) 2026-10-06 — The solution incorrectly calculates the function values at the grid points. Specifically, f(1/4) = 16/17 and f(3/4) = 16/25, but the equation in line 3 uses 16/25 for the term with weight 4 (corresponding to x=1/4) and 16/17 for the term with weight 4 (corresponding to x=3/4), effectively swapping the values or misidentifying the points. Additionally, the weight for the first term f(0)=1 is listed as 1/2 in the equation, whereas Simpson's rule requires a weight of 1 (multiplied by Δx/3 = 1/12). The correct sum should be (1*1 + 4*(16/17) + 2*(4/5) + 4*(16/25) + 1*1/2)/12.gpt-oss:20b: fail (error) 2026-10-06 — The solution miscomputes the function values at the sample points: it uses 1/2 for f(0.25) and 16/17 for f(1), which are incorrect. Consequently the weighted sum and the resulting Simpson estimate are wrong.
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 2026-10-06 with SymPy 1.14.0.