Trapezoidal, midpoint and Simpson's rules
Problem 4.388 · medium
Approximate \( \displaystyle \int_{0}^{2} \frac{x}{x + 1}\, dx \) using the Simpson's rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The Simpson's rule uses the points x = 0, 1/3, 2/3, 1, 4/3, 5/3, 2 with weights 1, 4, 2, 4, 2, 4, 1, all times 1/9.
- \[ \frac{1 \cdot 0 + 1 \cdot 2 \cdot \frac{1}{3} + 2 \cdot 2 \cdot \frac{1}{5} + 4 \cdot 1 \cdot \frac{1}{4} + 2 \cdot 4 \cdot \frac{1}{7} + 4 \cdot 1 \cdot \frac{1}{2} + 4 \cdot 5 \cdot \frac{1}{8}}{9} = \frac{1703}{1890} \]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 3.30e-04.✓ Proved
Answer \( S_{6} = \frac{1703}{1890} \approx 0.901058 \)
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 weighted sum in step 3 mis‑applies the function values: it uses 1·(2/3) instead of 1·f(1/3)=1·(1/4) and omits f(2)=2/3. The resulting Simpson estimate is incorrect.qwen3.6:27b-mlx: fail (error) — The solution incorrectly applies the multiplier for Simpson's Rule. The formula requires multiplying the weighted sum by \Delta x / 3 (which is 1/9), but the text states 'all times 1/9' while listing the standard weights 1, 4, 2... which already assume the factor of 1/3 is separate or the weights are 1/3, 4/3, etc. More critically, the calculation in line 3 divides by 9, but the weights listed (1,4,2...) are for the sum \sum w_i f(x_i) where the final result is (\Delta x / 3) * \sum. Here \Delta x = 1/3, so the multiplier is 1/9. However, the weights listed are 1, 4, 2, 4, 2, 4, 1. The standard Simpson's rule is (\Delta x / 3) * [f(x_0) + 4f(x_1) + 2f(x_2) + ... + f(x_n)]. With \Delta x = 1/3, the pre-factor is 1/9. The text says 'weights ... all times 1/9'. This is ambiguous but technically the calculation in line 3 is correct: it sums the weighted values and divides by 9. Wait, let's re-read carefully. Line 2 says 'weights 1, 4, 2, 4, 2, 4, 1, all times 1/9'. This implies the term is (1/9) * (1*f0 + 4*f1 + ...). This is correct. \Delta x = 1/3. Factor is \Delta x / 3 = 1/9. So the arithmetic is correct. Why did I think it was wrong? Let me check the function values. f(x) = x/(x+1). x0=0, f0=0. x1=1/3, f1=(1/3)/(4/3)=1/4. x2=2/3, f2=(2/3)/(5/3)=2/5. x3=1, f3=1/2. x4=4/3, f4=(4/3)/(7/3)=4/7. x5=5/3, f5=(5/3)/(8/3)=5/8. x6=2, f6=2/3. Sum = 1*0 + 4*(1/4) + 2*(2/5) + 4*(1/2) + 2*(4/7) + 4*(5/8) + 1*(2/3). Line 3 has: 1*0 + 1*(2/3) + 2*(2/5) + 4*(1/4) + 2*(4/7) + 4*(1/2) + 4*(5/8). The weights are permuted/wrong. The last term should be 1*f6 = 1*(2/3). Line 3 has 1*(2/3) as the second term? No, it has 1*(2/3) as the second term in the sum, but the second term corresponds to x1? No, the order in the sum in line 3 is: 1*0 (x0), 1*(2/3) (this is f6, but placed at index 1?), 2*(2/5) (f2), 4*(1/4) (f1), 2*(4/7) (f4), 4*(1/2) (f3), 4*(5/8) (f5). The weights are completely scrambled. The weight for f6 (x=2) should be 1. The weight for f1 (x=1/3) should be 4. In line 3, f6 (2/3) is multiplied by 1, but it is listed second. f1 (1/4) is multiplied by 4, but listed fourth. The sum is commutative, so the value might be correct if the pairs match. Let's check the pairs. Correct: 1*f0, 4*f1, 2*f2, 4*f3, 2*f4, 4*f5, 1*f6. Line 3: 1*f0, 1*f6, 2*f2, 4*f1, 2*f4, 4*f3, 4*f5. The weight for f5 is 4 in line 3. Correct weight for f5 is 4. The weight for f3 is 4 in line 3. Correct weight for f3 is 4. The weight for f1 is 4 in line 3. Correct weight for f1 is 4. The weight for f2 is 2 in line 3. Correct weight for f2 is 2. The weight for f4 is 2 in line 3. Correct weight for f4 is 2. The weight for f6 is 1 in line 3. Correct weight for f6 is 1. The weight for f0 is 1 in line 3. Correct weight for f0 is 1. So the set of products is correct, just ordered weirdly. The calculation is correct. Is there an error? Line 2 says 'weights 1, 4, 2, 4, 2, 4, 1'. This is the correct sequence of weights for n=6. Line 3 calculates the sum. The result 1703/1890 is correct. The exact value is correct. The approximation is correct. Why would it fail? Maybe the phrasing 'all times 1/9' is confusing? It says 'weights ... all times 1/9'. This usually means the weights are 1/9, 4/9, etc. But then it lists 1, 4, 2... This is a contradiction in description. It lists integer weights but says they are times 1/9. It should say 'multiplied by \Delta x / 3 = 1/9'. As written, it implies the weights themselves are 1/9, 4/9... but lists 1, 4... This is misleading/confusing. However, the calculation follows the standard formula. Let's look closer. 'weights 1, 4, 2, 4, 2, 4, 1, all times 1/9'. This is syntactically ambiguous. Does it mean (1, 4, ...) * 1/9? Or does it mean the weights are 1, 4... and the whole thing is times 1/9? Given line 3 divides by 9, it interprets it as the latter. But the sentence is poorly constructed. Is it an error? It's misleading. But wait, look at the weights in line 3 again. 1*0 + 1*(2/3) + ... The term 1*(2/3) is f(x_6). The term 4*(1/4) is f(x_1). The term 4*(1/2) is f(x_3). The term 4*(5/8) is f(x_5). The term 2*(2/5) is f(x_2). The term 2*(4/7) is f(x_4). The weights are correct for the respective function values. The order is just mixed up. The sentence in line 2 lists the weights in order 1, 4, 2, 4, 2, 4, 1. This corresponds to x_0 to x_6. Line 3 does not follow this order. This is not an error in math, just messy. Is there a deeper error? Simpson's rule requires n to be even. n=6 is even. Bounds 0 to 2. \Delta x = 2/6 = 1/3. Correct. The answer is correct. The only issue is the clumsy phrasing in line 2. 'weights ... all times 1/9'. If a student reads this, they might think the weights are 1/9, 4/9, 2/9... and then multiply by \Delta x again? No, line 3 clarifies the operation. I will mark it as misleading due to the ambiguous phrasing of the weights/multiplier relationship.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly applies the multiplier for Simpson's Rule. The formula requires multiplying the weighted sum by \Delta x / 3 (which is 1/9), but the text states 'all times 1/9' while listing the standard weights 1, 4, 2... which already assume the factor of 1/3 is separate or the weights are 1/3, 4/3, etc. More critically, the calculation in line 3 divides by 9, but the weights listed (1,4,2...) are for the sum \sum w_i f(x_i) where the final result is (\Delta x / 3) * \sum. Here \Delta x = 1/3, so the multiplier is 1/9. However, the weights listed are 1, 4, 2, 4, 2, 4, 1. The standard Simpson's rule is (\Delta x / 3) * [f(x_0) + 4f(x_1) + 2f(x_2) + ... + f(x_n)]. With \Delta x = 1/3, the pre-factor is 1/9. The text says 'weights ... all times 1/9'. This is ambiguous but technically the calculation in line 3 is correct: it sums the weighted values and divides by 9. Wait, let's re-read carefully. Line 2 says 'weights 1, 4, 2, 4, 2, 4, 1, all times 1/9'. This implies the term is (1/9) * (1*f0 + 4*f1 + ...). This is correct. \Delta x = 1/3. Factor is \Delta x / 3 = 1/9. So the arithmetic is correct. Why did I think it was wrong? Let me check the function values. f(x) = x/(x+1). x0=0, f0=0. x1=1/3, f1=(1/3)/(4/3)=1/4. x2=2/3, f2=(2/3)/(5/3)=2/5. x3=1, f3=1/2. x4=4/3, f4=(4/3)/(7/3)=4/7. x5=5/3, f5=(5/3)/(8/3)=5/8. x6=2, f6=2/3. Sum = 1*0 + 4*(1/4) + 2*(2/5) + 4*(1/2) + 2*(4/7) + 4*(5/8) + 1*(2/3). Line 3 has: 1*0 + 1*(2/3) + 2*(2/5) + 4*(1/4) + 2*(4/7) + 4*(1/2) + 4*(5/8). The weights are permuted/wrong. The last term should be 1*f6 = 1*(2/3). Line 3 has 1*(2/3) as the second term? No, it has 1*(2/3) as the second term in the sum, but the second term corresponds to x1? No, the order in the sum in line 3 is: 1*0 (x0), 1*(2/3) (this is f6, but placed at index 1?), 2*(2/5) (f2), 4*(1/4) (f1), 2*(4/7) (f4), 4*(1/2) (f3), 4*(5/8) (f5). The weights are completely scrambled. The weight for f6 (x=2) should be 1. The weight for f1 (x=1/3) should be 4. In line 3, f6 (2/3) is multiplied by 1, but it is listed second. f1 (1/4) is multiplied by 4, but listed fourth. The sum is commutative, so the value might be correct if the pairs match. Let's check the pairs. Correct: 1*f0, 4*f1, 2*f2, 4*f3, 2*f4, 4*f5, 1*f6. Line 3: 1*f0, 1*f6, 2*f2, 4*f1, 2*f4, 4*f3, 4*f5. The weight for f5 is 4 in line 3. Correct weight for f5 is 4. The weight for f3 is 4 in line 3. Correct weight for f3 is 4. The weight for f1 is 4 in line 3. Correct weight for f1 is 4. The weight for f2 is 2 in line 3. Correct weight for f2 is 2. The weight for f4 is 2 in line 3. Correct weight for f4 is 2. The weight for f6 is 1 in line 3. Correct weight for f6 is 1. The weight for f0 is 1 in line 3. Correct weight for f0 is 1. So the set of products is correct, just ordered weirdly. The calculation is correct. Is there an error? Line 2 says 'weights 1, 4, 2, 4, 2, 4, 1'. This is the correct sequence of weights for n=6. Line 3 calculates the sum. The result 1703/1890 is correct. The exact value is correct. The approximation is correct. Why would it fail? Maybe the phrasing 'all times 1/9' is confusing? It says 'weights ... all times 1/9'. This usually means the weights are 1/9, 4/9, etc. But then it lists 1, 4, 2... This is a contradiction in description. It lists integer weights but says they are times 1/9. It should say 'multiplied by \Delta x / 3 = 1/9'. As written, it implies the weights themselves are 1/9, 4/9... but lists 1, 4... This is misleading/confusing. However, the calculation follows the standard formula. Let's look closer. 'weights 1, 4, 2, 4, 2, 4, 1, all times 1/9'. This is syntactically ambiguous. Does it mean (1, 4, ...) * 1/9? Or does it mean the weights are 1, 4... and the whole thing is times 1/9? Given line 3 divides by 9, it interprets it as the latter. But the sentence is poorly constructed. Is it an error? It's misleading. But wait, look at the weights in line 3 again. 1*0 + 1*(2/3) + ... The term 1*(2/3) is f(x_6). The term 4*(1/4) is f(x_1). The term 4*(1/2) is f(x_3). The term 4*(5/8) is f(x_5). The term 2*(2/5) is f(x_2). The term 2*(4/7) is f(x_4). The weights are correct for the respective function values. The order is just mixed up. The sentence in line 2 lists the weights in order 1, 4, 2, 4, 2, 4, 1. This corresponds to x_0 to x_6. Line 3 does not follow this order. This is not an error in math, just messy. Is there a deeper error? Simpson's rule requires n to be even. n=6 is even. Bounds 0 to 2. \Delta x = 2/6 = 1/3. Correct. The answer is correct. The only issue is the clumsy phrasing in line 2. 'weights ... all times 1/9'. If a student reads this, they might think the weights are 1/9, 4/9, 2/9... and then multiply by \Delta x again? No, line 3 clarifies the operation. I will mark it as misleading due to the ambiguous phrasing of the weights/multiplier relationship.gpt-oss:20b: fail (error) 2026-10-05 — The weighted sum in step 3 mis‑applies the function values: it uses 1·(2/3) instead of 1·f(1/3)=1·(1/4) and omits f(2)=2/3. The resulting Simpson estimate is incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution incorrectly states that the weights are multiplied by 1/9; for n=6, Δx = 1/3, so the multiplier should be Δx/3 = 1/9, but the weights listed (1, 4, 2, 4, 2, 4, 1) are correct for Simpson's rule, implying the factor is indeed 1/9. However, the calculation in line 3 is wrong: the function values are evaluated incorrectly (e.g., f(1/3) = (1/3)/(4/3) = 1/4, not 2/3; f(2/3) = (2/3)/(5/3) = 2/5, etc., but the terms in the sum are mismatched with the x-values or weights). Specifically, the term for x=0 is 0, x=1/3 is 1/4, x=2/3 is 2/5, x=1 is 1/2, x=4/3 is 4/7, x=5/3 is 5/8, x=2 is 2/3. The sum in line 3 uses values like 2/3 for the first non-zero term which corresponds to x=2, not x=1/3. The arithmetic is garbled.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.