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Trapezoidal, midpoint and Simpson's rules

Problem 4.683 · medium

Approximate \( \displaystyle \int_{0}^{2} \frac{x}{x + 1}\, dx \) using the trapezoidal rule with \( \displaystyle n = 5 \).

\[ \frac{2}{5} \]
Δx = (b − a)/n.✓ Proved
The trapezoidal rule uses the points x = 0, 2/5, 4/5, 6/5, 8/5, 2 with weights 1, 2, 2, 2, 2, 1, all times 1/5.
\[ \frac{1 \cdot 0 + 2 \cdot 2 \cdot \frac{1}{7} + 1 \cdot 2 \cdot \frac{1}{3} + 2 \cdot 4 \cdot \frac{1}{9} + 2 \cdot 6 \cdot \frac{1}{11} + 2 \cdot 8 \cdot \frac{1}{13}}{5} = \frac{40078}{45045} \]
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 1.17e-02.✓ Proved

Answer \( T_{5} = \frac{40078}{45045} \approx 0.889732 \)

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: pass
qwen3.6:27b-mlx: fail (error) — The solution incorrectly applies the trapezoidal rule formula by dividing the weighted sum by 5 instead of 2*5 (or multiplying by Δx/2). The standard formula is T_n = (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)], but the solution uses (1/5) * [...], effectively calculating 2 * T_5.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies the trapezoidal rule formula by dividing the weighted sum by 5 instead of 2*5 (or multiplying by Δx/2). The standard formula is T_n = (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)], but the solution uses (1/5) * [...], effectively calculating 2 * T_5.
gpt-oss:20b: pass 2026-10-08
qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states that the weights are multiplied by 1/5. The standard trapezoidal rule formula is (Δx/2) * [f(x_0) + 2f(x_1) + ... +
gpt-oss:20b: pass 2026-10-08

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 2026-10-08 with SymPy 1.14.0.