Trapezoidal, midpoint and Simpson's rules
Problem 4.523 · medium
Approximate \( \displaystyle \int_{0}^{1} \frac{1}{x^{2} + 1}\, dx \) using the midpoint rule with \( \displaystyle n = 4 \).
- \[ 1 \cdot \frac{1}{4} = \frac{1}{4} \]Δx = (b − a)/n.✓ Proved
- The midpoint rule uses the points x = 1/8, 3/8, 5/8, 7/8 with weights 1, 1, 1, 1, all times 1/4.
- \[ \frac{1 \cdot 64 \cdot \frac{1}{113} + 1 \cdot 64 \cdot \frac{1}{89} + 1 \cdot 64 \cdot \frac{1}{73} + 1 \cdot 64 \cdot \frac{1}{65}}{4} = \frac{37541696}{47720465} \]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 1.30e-03.✓ Proved
Answer \( M_{4} = \frac{37541696}{47720465} \approx 0.786700 \)
Lines: 3 proved, 1 not checked. 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 | 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 is incorrect; the correct midpoint approximation is (64/65+64/73+64/89+64/113)/4≈0.8142, not 0.7867.qwen3.6:27b-mlx: pass — The solution correctly identifies the subinterval width and midpoints, applies the midpoint rule formula accurately, and provides the correct numerical approximation.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The weighted sum in step 3 is incorrect; the correct midpoint approximation is (64/65+64/73+64/89+64/113)/4≈0.8142, not 0.7867.qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies the subinterval width and midpoints, applies the midpoint rule formula accurately, and provides the correct numerical approximation.qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-06
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.