Trapezoidal, midpoint and Simpson's rules
Problem 4.521 · medium
Approximate \( \displaystyle \int_{1}^{2} \frac{1}{x}\, dx \) using the trapezoidal rule with \( \displaystyle n = 5 \).
- \[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 1, 6/5, 7/5, 8/5, 9/5, 2 with weights 1, 2, 2, 2, 2, 1, all times 1/10.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{2} + 1 \cdot 1 + 2 \cdot 5 \cdot \frac{1}{9} + 2 \cdot 5 \cdot \frac{1}{8} + 2 \cdot 5 \cdot \frac{1}{7} + 2 \cdot 5 \cdot \frac{1}{6}}{10} = \frac{1753}{2520} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{1}^{2} \frac{1}{x}\, dx = \ln{\left(2 \right)} \]The exact value, ≈ 0.693147, for comparison: the error is about 2.49e-03.✓ Proved
Answer \( T_{5} = \frac{1753}{2520} \approx 0.695635 \)
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 sentence describing the trapezoidal weights is incorrect: the end points should be weighted by Δx/2=0.05, not 0.1, and the interior points by Δx=0.1. Consequently the weighted sum in step 3 is computed with the wrong coefficients, leading to an inaccurate result.qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states the multiplier for the trapezoidal rule is 1/10 instead of Δx/2 = 1/10. While the final numerical result in step 3 i
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-06 — The sentence describing the trapezoidal weights is incorrect: the end points should be weighted by Δx/2=0.05, not 0.1, and the interior points by Δx=0.1. Consequently the weighted sum in step 3 is computed with the wrong coefficients, leading to an inaccurate result.qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states the multiplier for the trapezoidal rule is 1/10 instead of Δx/2 = 1/10. While the final numerical result in step 3 iqwen3.6:27b-mlx: fail (style) 2026-10-06 — [domain objection, downgraded to style] The solution incorrectly states the multiplier is 1/10, whereas the trapezoidal rule requires a factor of Δx/2 = 1/10 only if the weights are defined differently, but here the weights 1, 2, ..., 1 are used with a factor of Δx/2. Wait, standard formula is (Δx/2) * [f(x0) + 2f(x1) + ... + f(xn)]. Here Δx = 1/5, so Δx/2 = 1/10. The sentence says 'all times 1/10'. This is actually correct for the standard weights 1, 2, 2, 2, 2, 1. Let's re-read carefully. Sentence 2 says 'weights 1, 2, 2, 2, 2, 1, all times 1/10'. This implies the sum is (1/10) * (1*f(x0) + 2*f(x1) + ...). This is the correct formula. However, look at Equation 3. It calculates: (1*(1/2) + 1*1 + 2*(5/9) + ...)/10. The term 1*(1/2) corresponds to f(2) = 1/2. The term 1*1 corresponds to f(1) = 1. The terms 2*(5/9) etc correspond to 2*f(xi). So the equation implements the formula described in sentence 2. Is there an error? Let's check the points. x0=1, x1=1.2, x2=1.4, x3=1.6, x4=1.8, x5=2. f(1)=1. f(1.2)=5/6. f(1.4)=5/7. f(1.6)=5/8. f(1.8)=5/9. f(2)=1/2. The equation has: 1*(1/2) [f(2)] + 1*1 [f(1)] + 2*(5/9) [2f(1.8)] + 2*(5/8) [2f(1.6)] + 2*(5/7) [2f(1.4)] + 2*(5/6) [2f(1.2)]. This matches the points and weights. The factor is 1/10. This is correct. Why did I think it was wrong? Ah, I see. The sentence says 'weights 1, 2, 2, 2, 2, 1'. Usually, the 'weights' in the context of the sum Σ w_i f(x_i) are just 1, 2, 2, 2, 2, 1, and the whole thing is multiplied by Δx/2. The sentence says 'all times 1/10'. This is slightly ambiguous but mathematically correct if interpreted as the final multiplier. However, look at the order in Equation 3. It lists f(2) first, then f(1), then the interior points in reverse order (1.8, 1.6, 1.4, 1.2). This is fine for addition. Is there a subtle error? Let's check the calculation. 1/2 + 1 + 10/9 + 10/8 + 10/7 + 10/6. Sum = 1.5 + 1.111... + 1.25 + 1.428... + 1.666... = 6.956... Divide by 10 = 0.6956... The stated answer is 1753/2520 ≈ 0.695635. 1753/2520 = 0.6956349... The calculation is correct. Wait, is the sentence misleading? 'The trapezoidal rule uses the points ... with weights 1, 2, 2, 2, 2, 1, all times 1/10.' This is a correct description of the computation. Is there any other issue? The problem asks to approximate. The solution provides the approximation. The steps are correct. Why would it fail? Maybe the phrasing 'weights ... all times 1/10' is considered poor style because weights usually refer to the coefficients inside the sum, and the Δx/2 is a separate factor? Or is it an error? If I say 'weights are 1, 2, 2, 2, 2, 1', the sum is Σ w_i f(x_i). Then T_n = (Δx/2) Σ w_i f(x_i). The sentence says 'weights ... all times 1/10'. This could be interpreted as the weights themselves are 1/10, 2/10, etc. If the weights are 1/10, 2/10, etc, then the sum is Σ (w_i/10) f(x_i). This is equivalent to (1/10) Σ w_i f(x_i). So it is mathematically sound. However, standard terminology defines the weights as 1, 2, ..., 2, 1 and the factor as h/2. Combining them into 'weights ... times 1/10' is non-standard but not false. Let's look closer. Is there a mistake in the points? 1, 6/5, 7/5, 8/5, 9/5, 2. Correct. Weights 1, 2, 2, 2, 2, 1. Correct. Factor 1/10. Correct. Equation 3 matches. Equation 4 is just comparison. It seems correct. Did I miss something? 'all times 1/10' might be interpreted as multiplying the weights by 1/10, resulting in weights 0.1, 0.2, 0.2, 0.2, 0.2, 0.1. This is a valid way to describe the quadrature weights. I will mark it as pass unless I find a specific error. Wait, looking at Equation 3 again: `(1*(1/2) + 1*1 + 2*(5/9) + 2*(5/8) + 2*(5/7) + 2*(5/6))/10`. The terms are f(2), f(1), 2f(1.8), 2f(1.6), 2f(1.4), 2f(1.2). The sentence lists points x=1, 6/5, 7/5, 8/5, 9/5, 2. The weights 1, 2, 2, 2, 2, 1 correspond to these points in order. So weight 1 for x=1, weight 2 for x=6/5, etc. In Equation 3, the term `1*1` is for x=1 (weight 1). The term `2*(5/6)` is for x=6/5 (weight 2). The term `2*(5/7)` is for x=7/5 (weight 2). The term `2*(5/8)` is for x=8/5 (weight 2). The term `2*(5/9)` is for x=9/5 (weight 2). The term `1*(1/2)` is for x=2 (weight 1). The equation sums these up and divides by 10. This is correct. I see no error. Pass.gpt-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.