Trapezoidal, midpoint and Simpson's rules
Problem 4.826 · medium
Approximate \( \displaystyle \int_{0}^{2} \frac{x}{x + 1}\, dx \) 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.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{7} + 1 \cdot 1 \cdot \frac{1}{3} + 1 \cdot 5 \cdot \frac{1}{11} + 1 \cdot 7 \cdot \frac{1}{13} + 1 \cdot 3 \cdot \frac{1}{5} + 1 \cdot 11 \cdot \frac{1}{17}}{3} = \frac{693338}{765765} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{0}^{2} \frac{x}{x + 1}\, dx = 2 - \ln{\left(3 \right)} \]The exact value, ≈ 0.901388, for comparison: the error is about 4.03e-03.✓ Proved
Answer \( M_{6} = \frac{693338}{765765} \approx 0.905419 \)
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: passqwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-10qwen3.6:27b-mlx: inconclusive 2026-10-10 — reviewer returned a non-objectgpt-oss:20b: fail (error) 2026-10-10 — The weighted sum of the function values was computed incorrectly; the correct numerator is 693438, not 693338, leading to a different midpoint estimate.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the subinterval width, the midpoints, and the function evaluations. The final arithmetic and comparison with the exact value are correct.
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.