Trapezoidal, midpoint and Simpson's rules
Problem 4.678 · medium
Approximate \( \displaystyle \int_{1}^{3} \frac{1}{x}\, dx \) using the Simpson's rule with \( \displaystyle n = 4 \).
- \[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]Δx = (b − a)/n.✓ Proved
- The Simpson's rule uses the points x = 1, 3/2, 2, 5/2, 3 with weights 1, 4, 2, 4, 1, all times 1/6.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{3} + 1 \cdot 1 + 2 \cdot 1 \cdot \frac{1}{2} + 4 \cdot 2 \cdot \frac{1}{5} + 4 \cdot 2 \cdot \frac{1}{3}}{6} = \frac{11}{10} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{1}^{3} \frac{1}{x}\, dx = \ln{\left(3 \right)} \]The exact value, ≈ 1.098612, for comparison: the error is about 1.39e-03.✓ Proved
Answer \( S_{4} = \frac{11}{10} \approx 1.100000 \)
Lines: 3 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. A reviewer's objection to the explanation stands; see 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 weighted sum in step 3 is incorrect: the function values and Simpson weights are mismatched, leading to the wrong result 11/10. The correct computation gives 0.9333, not 1.1.qwen3.6:27b-mlx: fail (error) — The solution incorrectly applies Simpson's rule weights to the function values. The weights 1, 4, 2, 4, 1 correspond to the coefficients in the standard formula S_n = (Δx/3) * [f(x_0) + 4f(x_1) + 2f(x_2) + ... + f(x_n)]. The solution divides by 6 instead of 3 (or equivalently, uses Δx/6 instead of Δx/3). Since Δx = 1/2, the multiplier should be (1/2)/3 = 1/6, but the weights listed (1,4,2,4,1) are for the sum inside the bracket which is then multiplied by Δx/3. Wait, let's re-read carefully. Standard formula: S_n = (Δx/3) * Σ w_i f(x_i). Here Δx = 0.5. So multiplier is 0.5/3 = 1/6. The sentence says 'weights 1, 4, 2, 4, 1, all times 1/6'. This implies the calculation is (1/6) * (1*f0 + 4*f1 + 2*f2 + 4*f3 + 1*f4). This is actually correct arithmetic for Simpson's rule with n=4. Let's check the equation line 3. It calculates (1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3))/6. Wait, the function is 1/x. x values are 1, 1.5, 2, 2.5, 3. f(1)=1. f(1.5)=2/3. f(2)=1/2. f(2.5)=2/5. f(3)=1/3. The equation in line 3 has terms: 1*(1/3) [this is f(3)], 1*1 [this is f(1)], 2*(1/2) [this is f(2)], 4*(2/5) [this is f(2.5)], 4*(2/3) [this is f(1.5)]. The weights are applied to the wrong function values or the order is mixed up in the text description vs the equation. Sentence 2 says points are 1, 3/2, 2, 5/2, 3 with weights 1, 4, 2, 4, 1. This implies 1*f(1) + 4*f(3/2) + 2*f(2) + 4*f(5/2) + 1*f(3). Equation 3 calculates: 1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3). Let's map: 1/3 is f(3). 1 is f(1). 1/2 is f(2). 2/5 is f(5/2). 2/3 is f(3/2). So the equation computes: 1*f(3) + 1*f(1) + 2*f(2) + 4*f(5/2) + 4*f(3/2). The weights in the equation are 1, 1, 2, 4, 4. The weights in sentence 2 are 1, 4, 2, 4, 1. The equation applies weight 1 to f(1) and f(3), weight 2 to f(2), weight 4 to f(1.5) and f(2.5). This matches the standard Simpson's weights (1, 4, 2, 4, 1) applied to the correct points. The equation just lists them in a scrambled order (f(3), f(1), f(2), f(2.5), f(1.5)). The sum is commutative, so the result 11/10 is correct. Is there an error? Let's re-read sentence 2. 'weights 1, 4, 2, 4, 1'. This is the standard sequence. Equation 3 uses these weights on the correct values, just written in a weird order. The result is correct. Why did I think it was wrong? Ah, look at the equation again: `1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3)`. The coefficients are 1, 1, 2, 4, 4. The values are 1/3, 1, 1/2, 2/5, 2/3. f(1)=1, f(1.5)=2/3, f(2)=1/2, f(2.5)=2/5, f(3)=1/3. Correct weights for Simpson's are 1, 4, 2, 4, 1. So we need 1*f(1) + 4*f(1.5) + 2*f(2) + 4*f(2.5) + 1*f(3). The equation has 1*f(3) + 1*f(1) + 2*f(2) + 4*f(2.5) + 4*f(1.5). This is exactly the correct sum. The verdict should be pass? Wait, look at the sentence 2 again. It says 'weights 1, 4, 2, 4, 1'. It does not explicitly state they are multiplied by Δx/3. It says 'all times 1/6'. Since Δx = 1/2, Δx/3 = 1/6. This is correct. Is there a subtle error? The problem asks for an approximation. The answer is 1.1. The exact is ln(3) ≈ 1.0986. The approximation is reasonable. I don't see an error. Let me double check the 'adversarial' nature. Maybe the sentence 'The Simpson's rule uses the points... with weights... all times 1/6' is slightly ambiguous or non-standard phrasing, but mathematically correct. However, usually one writes S_n = (Δx/3)[...]. Saying 'weights ... all times 1/6' is a valid way to describe the final operation. I will mark it as pass.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies Simpson's rule weights to the function values. The weights 1, 4, 2, 4, 1 correspond to the coefficients in the standard formula S_n = (Δx/3) * [f(x_0) + 4f(x_1) + 2f(x_2) + ... + f(x_n)]. The solution divides by 6 instead of 3 (or equivalently, uses Δx/6 instead of Δx/3). Since Δx = 1/2, the multiplier should be (1/2)/3 = 1/6, but the weights listed (1,4,2,4,1) are for the sum inside the bracket which is then multiplied by Δx/3. Wait, let's re-read carefully. Standard formula: S_n = (Δx/3) * Σ w_i f(x_i). Here Δx = 0.5. So multiplier is 0.5/3 = 1/6. The sentence says 'weights 1, 4, 2, 4, 1, all times 1/6'. This implies the calculation is (1/6) * (1*f0 + 4*f1 + 2*f2 + 4*f3 + 1*f4). This is actually correct arithmetic for Simpson's rule with n=4. Let's check the equation line 3. It calculates (1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3))/6. Wait, the function is 1/x. x values are 1, 1.5, 2, 2.5, 3. f(1)=1. f(1.5)=2/3. f(2)=1/2. f(2.5)=2/5. f(3)=1/3. The equation in line 3 has terms: 1*(1/3) [this is f(3)], 1*1 [this is f(1)], 2*(1/2) [this is f(2)], 4*(2/5) [this is f(2.5)], 4*(2/3) [this is f(1.5)]. The weights are applied to the wrong function values or the order is mixed up in the text description vs the equation. Sentence 2 says points are 1, 3/2, 2, 5/2, 3 with weights 1, 4, 2, 4, 1. This implies 1*f(1) + 4*f(3/2) + 2*f(2) + 4*f(5/2) + 1*f(3). Equation 3 calculates: 1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3). Let's map: 1/3 is f(3). 1 is f(1). 1/2 is f(2). 2/5 is f(5/2). 2/3 is f(3/2). So the equation computes: 1*f(3) + 1*f(1) + 2*f(2) + 4*f(5/2) + 4*f(3/2). The weights in the equation are 1, 1, 2, 4, 4. The weights in sentence 2 are 1, 4, 2, 4, 1. The equation applies weight 1 to f(1) and f(3), weight 2 to f(2), weight 4 to f(1.5) and f(2.5). This matches the standard Simpson's weights (1, 4, 2, 4, 1) applied to the correct points. The equation just lists them in a scrambled order (f(3), f(1), f(2), f(2.5), f(1.5)). The sum is commutative, so the result 11/10 is correct. Is there an error? Let's re-read sentence 2. 'weights 1, 4, 2, 4, 1'. This is the standard sequence. Equation 3 uses these weights on the correct values, just written in a weird order. The result is correct. Why did I think it was wrong? Ah, look at the equation again: `1*(1/3) + 1*1 + 2*(1/2) + 4*(2/5) + 4*(2/3)`. The coefficients are 1, 1, 2, 4, 4. The values are 1/3, 1, 1/2, 2/5, 2/3. f(1)=1, f(1.5)=2/3, f(2)=1/2, f(2.5)=2/5, f(3)=1/3. Correct weights for Simpson's are 1, 4, 2, 4, 1. So we need 1*f(1) + 4*f(1.5) + 2*f(2) + 4*f(2.5) + 1*f(3). The equation has 1*f(3) + 1*f(1) + 2*f(2) + 4*f(2.5) + 4*f(1.5). This is exactly the correct sum. The verdict should be pass? Wait, look at the sentence 2 again. It says 'weights 1, 4, 2, 4, 1'. It does not explicitly state they are multiplied by Δx/3. It says 'all times 1/6'. Since Δx = 1/2, Δx/3 = 1/6. This is correct. Is there a subtle error? The problem asks for an approximation. The answer is 1.1. The exact is ln(3) ≈ 1.0986. The approximation is reasonable. I don't see an error. Let me double check the 'adversarial' nature. Maybe the sentence 'The Simpson's rule uses the points... with weights... all times 1/6' is slightly ambiguous or non-standard phrasing, but mathematically correct. However, usually one writes S_n = (Δx/3)[...]. Saying 'weights ... all times 1/6' is a valid way to describe the final operation. I will mark it as pass.gpt-oss:20b: fail (error) 2026-10-08 — The weighted sum in step 3 is incorrect: the function values and Simpson weights are mismatched, leading to the wrong result 11/10. The correct computation gives 0.9333, not 1.1.qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies the weights in the Simpson's rule formula. The weights 1, 4, 2, 4, 1 should be multiplied by the function values f(x_i) in order, but the equation in line 3 pairs them incorrectly (e.g., it uses 1 for f(1) and 1 for f(3/2) instead of 4). Specifically, the term for x=3/2 should have weight 4, not 1, and the term for x=2 should have weight 2, not 1. The calculation shown does not match the stated weights or the correct application of Simpson's rule.gpt-oss:20b: inconclusive 2026-10-08 — reviewer response could not be parsed: {"verdict":"fail","severity":"error","notes":"The weighted sum in step 3 misassigns the Simpson weights and function values. The correct sum is \((1·1 + 4·(2/3) + 2·(1/2) + 4·(2/5) + 1·(1/3))/6 = 11/1
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.