Trapezoidal, midpoint and Simpson's rules
Problem 4.833 · medium
Approximate \( \displaystyle \int_{0}^{2} x^{3} + 1\, dx \) using the midpoint rule with \( \displaystyle n = 5 \).
- \[ \frac{2}{5} \]Δx = (b − a)/n.✓ Proved
- The midpoint rule uses the points x = 1/5, 3/5, 1, 7/5, 9/5 with weights 1, 1, 1, 1, 1, all times 2/5.
- \[ \frac{2 \left(1 \cdot 126 \cdot \frac{1}{125} + 1 \cdot 152 \cdot \frac{1}{125} + 1 \cdot 2 + 1 \cdot 468 \cdot \frac{1}{125} + 1 \cdot 854 \cdot \frac{1}{125}\right)}{5} = \frac{148}{25} \]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 8.00e-02.✓ Proved
Answer \( M_{5} = \frac{148}{25} \approx 5.920000 \)
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-objectqwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the subinterval width and midpoints, and the final calculation is accurate.gpt-oss:20b: fail (error) 2026-10-10 — The computed midpoint sum is incorrect: Δx=2/5 and the sum of f at the midpoints equals 1900/125=15.2, giving M₅=0.4·15.2=6.08, not 148/25=5.92. The solution’s arithmetic is 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 2026-10-10 with SymPy 1.14.0.