Trapezoidal, midpoint and Simpson's rules
Problem 4.831 · medium
Approximate \( \displaystyle \int_{0}^{1} \frac{4}{x^{2} + 1}\, dx \) using the trapezoidal rule with \( \displaystyle n = 6 \).
- \[ 1 \cdot \frac{1}{6} = \frac{1}{6} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 0, 1/6, 1/3, 1/2, 2/3, 5/6, 1 with weights 1, 2, 2, 2, 2, 2, 1, all times 1/12.
- \[ \frac{1 \cdot 2 + 1 \cdot 4 + 2 \cdot 144 \cdot \frac{1}{61} + 2 \cdot 36 \cdot \frac{1}{13} + 2 \cdot 16 \cdot \frac{1}{5} + 2 \cdot 18 \cdot \frac{1}{5} + 2 \cdot 144 \cdot \frac{1}{37}}{12} = \frac{2761249}{880230} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{0}^{1} \frac{4}{x^{2} + 1}\, dx = \pi \]The exact value, ≈ 3.141593, for comparison: the error is about 4.63e-03.✓ Proved
Answer \( T_{6} = \frac{2761249}{880230} \approx 3.136963 \)
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 solution lists incorrect function values (e.g., 144/61 instead of 144/37) and misapplies the trapezoidal weights, so the computed sum is not the correct trapezoidal approximation.qwen3.6:27b-mlx: pass — The solution correctly identifies the step size, grid points, and weights for the trapezoidal rule. The arithmetic in the weighted sum is correct, and the final approximation matches the stated answer.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-10 — The solution lists incorrect function values (e.g., 144/61 instead of 144/37) and misapplies the trapezoidal weights, so the computed sum is not the correct trapezoidal approximation.qwen3.6:27b-mlx: pass 2026-10-10 — The solution correctly identifies the step size, grid points, and weights for the trapezoidal rule. The arithmetic in the weighted sum is correct, and the final approximation matches the stated answer.qwen3.6:27b-mlx: fail (error) 2026-10-10 — The solution incorrectly states that the trapezoidal rule weights are multiplied by 1/12. The correct multiplier is Δx/2 = 1/12, but the sentence implies the weights themselves are scaled by 1/12 rather than the sum being multiplied by Δx/2, which is a confusing and potentially misleading description of the formula T_n = (Δx/2)[f(x_0) + 2f(x_1) + ... + f(x_n)]. More critically, the equation in line 3 divides by 12, which is correct for Δx/2, but the sentence in line 2 says 'all times 1/12' referring to the weights, which is imprecise phrasing for the standard formula structure. However, looking closer at line 3, the division by 12 is applied to the sum. The sentence says 'weights ... all times 1/12'. This is technically incorrect terminology; the weights are 1, 2, ..., 1 and the whole sum is multiplied by Δx/2. But is it an error? Yes, it misstates the algorithm. The weights are integers; the factor is Δx/2. Saying the weights are 'times 1/12' suggests the weights are 1/12, 2/12, etc., which is not the standard definition, though numerically equivalent if applied to the sum. Let's look for a harder error. Actually, let's re-read carefully. Line 2: 'weights 1, 2, 2, 2, 2, 2, 1, all times 1/12'. This implies the term is (1/12)*1*f(x0) + ... which is correct numerically. Let's check the function values in Line 3. f(x) = 4/(x^2+1). x0=0, f=4. Term: 1*4 = 4. Line 3 has 1*2. ERROR. f(0) = 4/(0+1) = 4. The line has 1*2. x1=1/6, f=4/(1/36+1) = 4/(37/36) = 144/37. Line 3 has 1*4? No, it has 1*4 for the second term? Wait. Line 3: (1*2 + 1*4 + ...). f(0) = 4. Weight 1. Term should be 4. Line 3 has 1*2 = 2. f(1) = 4/(1+1) = 2. Weight 1. Term should be 2. It seems the solver swapped f(0) and f(1) or just calculated f(0) wrong as 2? f(0) = 4. f(1) = 2. Line 3 starts with 1*2. This corresponds to f(1). Line 3 ends with 2*(144/37). 144/37 is f(1/6). The terms are mixed up or calculated incorrectly. f(0)=4. f(1/6)=144/37. f(1/3)=4/(1/9+1)=36/10=18/5. f(1/2)=4/(1/4+1)=16/5. f(2/3)=4/(4/9+1)=36/13. f(5/6)=144/37. f(1)=2. Correct Sum: 1*4 + 2*(144/37) + 2*(18/5) + 2*(16/5) + 2*(36/13) + 2*(144/37) + 1*2. Line 3 Sum: 1*2 + 1*4 + 2*(144/61) ... 144/61? f(1/6) is 144/37. 144/61 is wrong. There are multiple calculation errors in the equation line, but the prompt says 'equation ... checked by CAS ... do not re-check the algebra of a line marked proved'. Wait, the prompt says 'Each line ... is either an EQUATION ... (its result is shown; do not re-check the algebra of a line marked proved)'. Line 3 is marked [equation, unchecked]. The text says 'unchecked' but the instruction says 'do not re-check the algebra of a line marked proved'. Here it is marked 'unchecked'. Does 'unchecked' mean I should check it? Or does it mean the CAS didn't check it? The prompt says: 'Each line ... is either an EQUATION, which a computer algebra system has already checked (its result is shown; do not re-check the algebra of a line marked proved)'. Line 3 is marked [equation, unchecked]. This likely means the CAS did NOT check it, or the label is just a tag. But the instruction says 'do not re-check the algebra of a line marked proved'. Line 3 is NOT marked proved. It is marked unchecked. However, usually in these tasks, 'equation' lines are treated as atomic facts if they are marked 'proved'. If they are 'unchecked', I might need to verify them? Let's look at Line 2. 'The trapezoidal rule uses the points ... with weights ... all times 1/12'. This sentence describes the method. Is the description correct? T_n = (Δx/2) [f(x0) + 2f(x1) + ... + f(xn)]. Δx = 1/6. Δx/2 = 1/12. So T_n = (1/12) [1*f(x0) + 2*f(x1) + ... + 1*f(xn)]. The sentence says 'weights 1, 2, ..., 1, all times 1/12'. This can be interpreted as: The terms are (1/12)*1*f(x0), etc. This is numerically correct. But is it misleading? Standard weights are 1, 2, ..., 1. The factor is Δx/2. Saying the weights are 'all times 1/12' conflates the step size factor with the weights. But the bigger issue is likely the setup or the result. Let's look at the result. T_6 = 2761249/880230 ≈ 3.136963. Exact = pi ≈ 3.14159. Error ≈ 0.0046. Let's calculate T_6 properly. f(0)=4. f(1/6)=144/37 ≈ 3.8919. f(1/3)=18/5 = 3.6. f(1/2)=16/5 = 3.2. f(2/3)=36/13 ≈ 2.7692. f(5/6)=144/37 ≈ 3.8919. f(1)=2. Sum = 4 + 2 + 2*(144/37 + 18/5 + 16/5 + 36/13 + 144/37). Inner sum = 288/37 + 34/5 + 36/13. 288/37 ≈ 7.7838. 34/5 = 6.8. 36/13 ≈ 2.7692. Sum ≈ 17.353. Total Sum = 6 + 2*17.353 = 6 + 34.706 = 40.706. T_6 = (1/12) * 40.706 ≈ 3.392. Wait. Let's re-evaluate f(x). f(x) = 4/(x^2+1). f(0) = 4. f(1) = 2. My manual calculation: Sum = 1*4 + 2*(144/37) + 2*(18/5) + 2*(16/5) + 2*(36/13) + 2*(144/37) + 1*2. = 6 + 2*(288/37 + 34/5 + 36/13). 288/37 = 7.78378. 34/5 = 6.8. 36/13 = 2.76923. Sum inside = 17.35301. 2 * Sum = 34.70602. Total = 40.70602. Divide by 12: 3.392168. The stated answer is 3.136963. These are very different. Why? Did the problem ask for Simpson's rule? No, Trapezoidal. Did I calculate f(x) wrong? f(1/2) = 4/(1/4+1) = 4/(5/4) = 16/5 = 3.2. Correct. f(1/3) = 4/(1/9+1) = 4/(10/9) = 36/10 = 3.6. Correct. f(2/3) = 4/(4/9+1) = 4/(13/9) = 36/13 ≈ 2.77. Correct. f(1/6) = 4/(1/36+1) = 4/(37/36) = 144/37 ≈ 3.89. Correct. So T_6 should be approx 3.39. The stated answer 3.137 is close to Pi (3.14159). Is it possible the solution used n=6 for Simpson's rule? S_6 = (Δx/3)[f(0) + 4f(1/6) + 2f(1/3) + 4f(1/2) + 2f(2/3) + 4f(5/6) + f(1)]. Δx/3 = 1/18. Sum = 4 + 2 + 4*(144/37 + 144/37) + 2*(18/5 + 36/13) + 4*(16/5). = 6 + 4*(288/37) + 2*(18/5 + 36/13) + 64/5. = 6 + 1152/37 + 36/5 + 72/13 + 64/5. = 6 + 31.135 + 7.2 + 5.538 + 12.8. = 62.673. S_6 = 62.673 / 18 ≈ 3.48. Still not 3.137. Let's look at Line 3 again. (1*2 + 1*4 + 2*(144/61) + 2*(36/13) + 2*(16/5) + 2*(18/5) + 2*(144/37))/12. Note the term 144/61. If x=1/6, x^2+1 = 37/36. 4/(37/36) = 144/37. Where does 61 come from? Maybe the function was different? Or maybe the points are different? If the solution is numerically wrong, and I am not supposed to check the algebra of 'proved' lines, but Line 3 is 'unchecked', I should probably flag the numerical discrepancy if it's a 'sentence' error? No, Line 3 is an equation. Line 2 is a sentence. Line 2 says: 'The trapezoidal rule uses the points ... with weights ... all times 1/12'. This sentence is the setup. If the setup is correct, but the execution (Line 3) is wrong, and Line 3 is unchecked, do I fail the solution? The prompt asks me to review the solution. If the final answer is wrong, the solution fails. The error is in Line 3 (calculation) or Line 2 (description)? Line 2's description is slightly loose ('weights ... times 1/12') but standard enough. Line 3 has '144/61'. This is a calculation error. Since Line 3 is marked 'unchecked', I am allowed/expected to check it? The prompt says: 'do not re-check the algebra of a line marked proved'. It does NOT say 'do not check unchecked lines'. Therefore, I must check Line 3. Line 3 contains 144/61. f(1/6) = 144/37. 144/61 is incorrect. Also, the first term is 1*2. f(0)=4. It should be 1*4. The last term is 2*(144/37). f(1)=2. It should be 1*2. The terms are scrambled and values are wrong. Thus, the solution is erroneous. Verdict: fail. Severity: error. Notes: The function values in the summation are incorrect (e.g., f(0)=4 is written as 2, and f(1/6)=144/37 is written as 144/61), leading to an incorrect final result.gpt-oss:20b: fail (misleading) 2026-10-10 — The sentence claims that all function values are multiplied by 1/12, but in the trapezoidal rule the interior points receive a weight of 2·(Δx/2)=Δx=1/6, while the endpoints receive Δx/2=1/12. This misstates the weighting scheme and would lead a student to apply the wrong coefficients.
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.