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Trapezoidal, midpoint and Simpson's rules

Problem 4.828 · medium

Approximate \( \displaystyle \int_{0}^{2} x^{3} + 1\, dx \) using the Simpson's rule with \( \displaystyle n = 4 \).
  1. \[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]
    Δx = (b − a)/n.✓ Proved
  2. The Simpson's rule uses the points x = 0, 1/2, 1, 3/2, 2 with weights 1, 4, 2, 4, 1, all times 1/6.
  3. \[ \frac{1 \cdot 1 + 2 \cdot 2 + 4 \cdot 9 \cdot \frac{1}{8} + 1 \cdot 9 + 4 \cdot 35 \cdot \frac{1}{8}}{6} = 6 \]
    Weighted sum of the function values.✓ Proved
  4. \[ \int\limits_{0}^{2} \left(x^{3} + 1\right)\, dx = 6 \]
    The exact value, ≈ 6.000000, for comparison: the error is about 0.00e+00.✓ Proved
Answer \( S_{4} = 6 \approx 6.000000 \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the rule implemented separately in floating point

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution incorrectly states the weights for Simpson's Rule as 'all times 1/6'. The correct multiplier is Δx/3, which is 1/6 in this specific case, but the general formula requires dividing by 3, not 6 (the weights 1, 4, 1 are multiplied by Δx/3). Stating the weights are 'times 1/6' implies the standard weights are 1/6, 4/6, etc., which is false; the weights are 1, 4, 2, 4, 1 multiplied by Δx/3. Additionally, the function values listed in step 3 do not match the points listed in step 2 (e.g., f(0)=1, f(0.5)=1.125, f(1)=2, f(1.5)=4.375, f(2)=9). The calculation in step 3 uses f(0)=1, f(0.5)=2 (wrong), f(1)=9/8 (wrong), etc. It seems to have swapped or miscalculated the function values entirely while coincidentally getting the right sum due to the specific polynomial degree, but the intermediate values are wrong.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-10
  • qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution incorrectly states the weights for Simpson's Rule as 'all times 1/6'. The correct multiplier is Δx/3, which is 1/6 in this specific case, but the general formula requires dividing by 3, not 6 (the weights 1, 4, 1 are multiplied by Δx/3). Stating the weights are 'times 1/6' implies the standard weights are 1/6, 4/6, etc., which is false; the weights are 1, 4, 2, 4, 1 multiplied by Δx/3. Additionally, the function values listed in step 3 do not match the points listed in step 2 (e.g., f(0)=1, f(0.5)=1.125, f(1)=2, f(1.5)=4.375, f(2)=9). The calculation in step 3 uses f(0)=1, f(0.5)=2 (wrong), f(1)=9/8 (wrong), etc. It seems to have swapped or miscalculated the function values entirely while coincidentally getting the right sum due to the specific polynomial degree, but the intermediate values are wrong.
  • qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution incorrectly states that the weights are multiplied by 1/6. For Simpson's Rule, the factor is Δx/3, which is (1/2)/3 = 1/6. However, the text says 'all times 1/6' implying the weights 1,4,2,4,1 are multiplied by 1/6 directly, but standard notation usually implies the sum is multiplied by Δx/3. Let's look closer. The equation in line 3 divides the sum by 6. The sum inside is 1*1 + 4*(1.25) + 2*(2) + 4*(4.375) + 1*(9)? No, let's check the values. f(0)=1, f(0.5)=1.125, f(1)=2, f(1.5)=4.375, f(2)=9. Weights 1,4,2,4,1. Sum = 1 + 4.5 + 4 + 17.5 + 9 = 36. 36 * (1/6) = 6. The calculation is correct. The sentence says 'weights 1, 4, 2, 4, 1, all times 1/6'. This is ambiguous but technically the multiplier for the whole sum is 1/6. Wait, standard Simpson's is (Δx/3) * [f0 + 4f1 + 2f2 + 4f3 + f4]. Here Δx/3 = 1/6. So the sentence is correct in its implication. Is there an error? Line 1 says '1/2 = 1/2'. This is a tautology, not a calculation of Δx. It should state Δx = 1/2. But it's marked 'unchecked equation'. The prompt says 'do not re-check the algebra of a line marked proved'. Line 1 is 'unchecked'. But it's an equation. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. Line 1 is an equation. I should not check its algebra. But is it a valid setup? It's just a number. The main issue is likely the phrasing in line 2 or the values in line 3. Let's re-read line 3. '(1*1 + 2*2 + 4*(9/8) + 1*9 + 4*(35/8))/6'. The terms are: 1*f(0) + 2*f(1) + 4*f(0.5) + 1*f(2) + 4*f(1.5). The order is mixed up (0, 1, 0.5, 2, 1.5) but addition is commutative. The values: f(0)=1. f(1)=2. f(0.5)=1.125=9/8. f(2)=9. f(1.5)=4.375=35/8. The weights applied: 1 to f(0), 2 to f(1), 4 to f(0.5), 1 to f(2), 4 to f(1.5). This matches the Simpson weights 1, 4, 2, 4, 1 assigned to x0, x1, x2, x3, x4. So the calculation is correct. Why would it fail? Maybe the sentence in line 2 is misleading? 'The Simpson's rule uses the points... with weights... all times 1/6'. This is a correct description of the formula S_n = (Δx/3) Σ w_i f(x_i). Here Δx/3 = 1/6. So it is correct. Is there a subtle error? The problem asks to approximate. The answer is 6. The exact integral is 6. The error is 0. This is correct for a cubic polynomial. Simpson's rule is exact for cubics. So the result is correct. The setup is correct. The sentences are correct. Why 'fail'? Maybe the 'unchecked' tag on line 1 is weird? No, I must accept equations. Maybe the sentence in line 2 is considered 'style' or 'misleading' because it doesn't explicitly state Δx? Or maybe the order of points in line 3 is confusing? The prompt asks to find what is wrong. If nothing is wrong, I say pass. But usually these adversarial reviews have a trap. Let's look at line 2 again. 'The Simpson's rule uses the points x = 0, 1/2, 1, 3/2, 2 with weights 1, 4, 2, 4, 1, all times 1/6.' This is a bit colloquial. 'All times 1/6' could mean each weight is multiplied by 1/6, which is correct. Or it could mean the sum is multiplied by 1/6. Both are mathematically equivalent. Is there a hypothesis violation? No. Is the region right? Yes. Is the direction right? Yes. I will bet on 'pass' unless I see a specific error. Wait, look at line 3 again. `1*1 + 2*2 + 4*(9/8) + 1*9 + 4*(35/8)`. The weights are 1, 2, 4, 1, 4. The standard weights are 1, 4, 2, 4, 1. The term `2*2` corresponds to weight 2 on f(1). The term `4*(9/8)` corresponds to weight 4 on f(0.5). The term `4*(35/8)` corresponds to weight 4 on f(1.5). The term `1*1` is weight 1 on f(0). The term `1*9` is weight 1 on f(2). The set of weights {1, 2, 4, 1, 4} is the same as {1, 4, 2, 4, 1}. The assignment is correct. The sum is correct. The divisor is 6. The result is 6. Everything seems correct. I will output pass.
  • gpt-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.

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