Trapezoidal, midpoint and Simpson's rules
Problem 4.830 · medium
Approximate \( \displaystyle \int_{0}^{1} \frac{4}{x^{2} + 1}\, dx \) using the trapezoidal rule with \( \displaystyle n = 5 \).
- \[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 0, 1/5, 2/5, 3/5, 4/5, 1 with weights 1, 2, 2, 2, 2, 1, all times 1/10.Reviewed
- \[ \frac{1 \cdot 2 + 1 \cdot 4 + 2 \cdot 100 \cdot \frac{1}{41} + 2 \cdot 50 \cdot \frac{1}{17} + 2 \cdot 100 \cdot \frac{1}{29} + 2 \cdot 50 \cdot \frac{1}{13}}{10} = \frac{4118807}{1313845} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{0}^{1} \frac{4}{x^{2} + 1}\, dx = \pi \]The exact value, ≈ 3.141593, for comparison: the error is about 6.67e-03.✓ Proved
Answer \( T_{5} = \frac{4118807}{1313845} \approx 3.134926 \)
✓ Nihil obstat Lines: 3 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
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 with n=5. The arithmetic in the weighted sum is correct, and the final approximation matches the stated answer.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the step size, grid points, and weights for the trapezoidal rule with n=5. The arithmetic in the weighted sum is correct, and the final approximation matches the stated answer.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the step size, the grid points, and the weights for the trapezoidal rule. The arithmetic in the weighted sum is correct, leading to the stated approximation.gpt-oss:20b: pass 2026-10-10
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.