Trapezoidal, midpoint and Simpson's rules
Problem 4.974 · medium
Approximate \( \displaystyle \int_{0}^{1} \frac{1}{x^{2} + 1}\, dx \) using the trapezoidal rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{6} = \frac{1}{6} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 0, 1/6, 1/3, 1/2, 2/3, 5/6, 1 with weights 1, 2, 2, 2, 2, 2, 1, all times 1/12.Reviewed
- \[ \frac{1 \cdot 1 \cdot \frac{1}{2} + 1 \cdot 1 + 2 \cdot 36 \cdot \frac{1}{61} + 2 \cdot 9 \cdot \frac{1}{13} + 2 \cdot 4 \cdot \frac{1}{5} + 2 \cdot 9 \cdot \frac{1}{10} + 2 \cdot 36 \cdot \frac{1}{37}}{12} = \frac{2761249}{3520920} \]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 1.16e-03.✓ Proved
Answer \( T_{6} = \frac{2761249}{3520920} \approx 0.784241 \)
Lines: 3 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly identifies the step size, grid points, and weights for the trapezoidal rule. The arithmetic is verified by the CAS, and the final approximation is correct.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: pass 2026-10-12 — The solution correctly identifies the step size, grid points, and weights for the trapezoidal rule. The arithmetic is verified by the CAS, and the final approximation is correct.gpt-oss:20b: pass 2026-10-12qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly lists the function values in the weighted sum. For example, it uses 1/2 for f(0) instead of 1, and 1 for f(1) instead of 1/2. Additionally, the value for f(1/2) is listed as 36/61 (which is f(5/6)) instead of 4/5. The final numerical result is coincidentally correct due to the symmetry of the function and the reversal of the endpoint weights, but the intermediate steps are mathematically 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 2026-10-12 with SymPy 1.14.0.