Trapezoidal, midpoint and Simpson's rules
Problem 4.973 · medium
Approximate \( \displaystyle \int_{1}^{2} \frac{1}{x^{2}}\, dx \) using the trapezoidal rule with \( \displaystyle n = 4 \).
- \[ 1 \cdot \frac{1}{4} = \frac{1}{4} \]Δx = (b − a)/n.✓ Proved
- The trapezoidal rule uses the points x = 1, 5/4, 3/2, 7/4, 2 with weights 1, 2, 2, 2, 1, all times 1/8.
- \[ \frac{1 \cdot 1 \cdot \frac{1}{4} + 2 \cdot 16 \cdot \frac{1}{49} + 2 \cdot 4 \cdot \frac{1}{9} + 1 \cdot 1 + 2 \cdot 16 \cdot \frac{1}{25}}{8} = \frac{179573}{352800} \]Weighted sum of the function values.✓ Proved
- \[ \int\limits_{1}^{2} \frac{1}{x^{2}}\, dx = \frac{1}{2} \]The exact value, ≈ 0.500000, for comparison: the error is about 8.99e-03.✓ Proved
Answer \( T_{4} = \frac{179573}{352800} \approx 0.508994 \)
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 solution misidentifies the function values: it uses 1/4 instead of f(1)=1 and mixes up the values for x=5/4 and x=7/4. The weighted sum is therefore incorrect.qwen3.6:27b-mlx: fail (error) — The solution incorrectly calculates the function values at the grid points. For example, f(5/4) is 16/25, not 16/49, and f(7/4) is 16/49, not 16/25. The weights are also incorrect; the trapezoidal rule requires multiplying the sum by Δx/2 = 1/8, but the sentence claims 'all times 1/8' while listing weights 1,2,2,2,1 which implies the standard formula T_n = (Δx/2)[f(x_0) + 2f(x_1) + ... + f(x_n)]. However, the equation in line 3 uses a divisor of 8, which is correct for Δx/2, but the function values inside are swapped/wrong. Specifically, f(1)=1, f(5/4)=16/25, f(3/2)=4/9, f(7/4)=16/49, f(2)=1/4. The solution lists 16/49 for the second term (should be 16/25) and 16/25 for the fourth term (should be 16/49). This is a calculation error in the setup.
Every verdict on record (4)
gpt-oss:20b: fail (error) 2026-10-12 — The solution misidentifies the function values: it uses 1/4 instead of f(1)=1 and mixes up the values for x=5/4 and x=7/4. The weighted sum is therefore incorrect.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly calculates the function values at the grid points. For example, f(5/4) is 16/25, not 16/49, and f(7/4) is 16/49, not 16/25. The weights are also incorrect; the trapezoidal rule requires multiplying the sum by Δx/2 = 1/8, but the sentence claims 'all times 1/8' while listing weights 1,2,2,2,1 which implies the standard formula T_n = (Δx/2)[f(x_0) + 2f(x_1) + ... + f(x_n)]. However, the equation in line 3 uses a divisor of 8, which is correct for Δx/2, but the function values inside are swapped/wrong. Specifically, f(1)=1, f(5/4)=16/25, f(3/2)=4/9, f(7/4)=16/49, f(2)=1/4. The solution lists 16/49 for the second term (should be 16/25) and 16/25 for the fourth term (should be 16/49). This is a calculation error in the setup.gpt-oss:20b: fail (error) 2026-10-12 — The solution misidentifies the function values at x=5/4 and x=7/4, using 16/49 for f(5/4) and 16/25 for f(7/4). This incorrect substitution leads to an erroneous trapezoidal estimate.qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly calculates the function values for the trapezoidal rule. For example, f(5/4) = (5/4)^(-2) = 16/25, but the solution uses 16/49 (which is f(7/4)). Similarly, it uses 16/25 for f(7/4) instead of 16/49. The weights and points are listed correctly, but the substitution into the formula is wrong.
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 2026-10-12 with SymPy 1.14.0.