Trapezoidal, midpoint and Simpson's rules
Problem 4.832 · medium
Approximate \( \displaystyle \int_{0}^{2} x^{3} + 1\, dx \) using the midpoint rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The midpoint rule uses the points x = 1/6, 1/2, 5/6, 7/6, 3/2, 11/6 with weights 1, 1, 1, 1, 1, 1, all times 1/3.Reviewed
- \[ \frac{1 \cdot 217 \cdot \frac{1}{216} + 1 \cdot 9 \cdot \frac{1}{8} + 1 \cdot 341 \cdot \frac{1}{216} + 1 \cdot 559 \cdot \frac{1}{216} + 1 \cdot 35 \cdot \frac{1}{8} + 1 \cdot 1547 \cdot \frac{1}{216}}{3} = \frac{107}{18} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{0}^{2} \left(x^{3} + 1\right)\, dx = 6 \]The exact value, ≈ 6.000000, for comparison: the error is about 5.56e-02.✓ Proved
Answer \( M_{6} = \frac{107}{18} \approx 5.944444 \)
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 subinterval width and midpoints, and the final calculation is accurate.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the subinterval width and midpoints, and the final calculation is accurate.qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer returned a non-objectgpt-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.