Home›Calculus 2›Trapezoidal, midpoint and Simpson's rules›Problem 4.527

Trapezoidal, midpoint and Simpson's rules

Problem 4.527 · medium

Approximate \( \displaystyle \int_{1}^{2} \frac{1}{x}\, dx \) 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.
\[ \frac{1 \cdot 1 \cdot \frac{1}{2} + 1 \cdot 1 + 2 \cdot 4 \cdot \frac{1}{7} + 2 \cdot 2 \cdot \frac{1}{3} + 2 \cdot 4 \cdot \frac{1}{5}}{8} = \frac{1171}{1680} \]
Weighted sum of the function values.✓ Proved
\[ \int\limits_{1}^{2} \frac{1}{x}\, dx = \ln{\left(2 \right)} \]
The exact value, ≈ 0.693147, for comparison: the error is about 3.88e-03.✓ Proved

Answer \( T_{4} = \frac{1171}{1680} \approx 0.697024 \)

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: pass
qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states that the weights are multiplied by 1/8. The trapezoidal rule formula is (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)].

Every verdict on record (4)

gpt-oss:20b: pass 2026-10-06
qwen3.6:27b-mlx: inconclusive 2026-10-06 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly states that the weights are multiplied by 1/8. The trapezoidal rule formula is (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)].
qwen3.6:27b-mlx: fail (style) 2026-10-06 — [domain objection, downgraded to style] The solution incorrectly states that the weights are multiplied by 1/8. The trapezoidal rule formula is (Δx/2) * [f(x0) + 2f(x1) + ... + f(xn)]. With Δx = 1/4, the factor should be (1/4)/2 = 1/8, but the weights listed (1, 2, 2, 2, 1) are already the coefficients inside the bracket. The sentence implies the entire sum is multiplied by 1/8, which is correct for the factor, but the phrasing 'weights ... all times 1/8' is ambiguous and potentially confusing regarding whether the weights are 1,2,2,2,1 or 1/8, 2/8, etc. More critically, looking at line 3, the calculation is (sum)/8. Let's check the sum: 1/2 + 1 + 8/7 + 4/3 + 8/5. Common denominator 105? No, 1680. 1/2 = 840/1680. 1 = 1680/1680. 8/7 = 1920/1680. 4/3 = 2240/1680. 8/5 = 2688/1680. Sum = 840+1680+1920+2240+2688 = 9368. 9368/1680 = 1171/210. Wait. The equation says the result is 1171/1680. 9368/8 = 1171. So the calculation in line 3 is (Sum)/8 = 1171/1680? No. (Sum)/8 = 9368/1680 = 1171/210 ≈ 5.57. The correct T4 is (1/8) * (1/2 + 1 + 8/7 + 4/3 + 8/5) = 1/8 * (9368/1680) = 1171/1680. Ah, the equation in line 3 is `(Sum)/8 = 1171/1680`. Let's re-evaluate. Sum = 9368/1680. Divided by 8 is 1171/1680. So the math in line 3 is correct. The issue is the description in line 2. It says 'weights 1, 2, 2, 2, 1, all times 1/8'. This usually means the weights are 1/8, 2/8, 2/8, 2/8, 1/8. If you multiply f(x) by these weights, you get the correct answer. However, standard terminology defines weights as the coefficients 1, 2, ..., 2, 1, and the factor Δx/2 is separate. The sentence is slightly ambiguous but not strictly 'wrong' if interpreted as 'the terms are weighted by 1, 2, 2, 2, 1 and the whole thing is multiplied by 1/8'. But wait, look at line 3 again. `(1*(1/2) + 1*1 + 2*(4/7) + 2*(2/3) + 2*(4/5))/8`. The terms inside are `1*f(1)`, `1*f(2)`? No. `f(1)=1`. `f(5/4)=4/5`. `f(3/2)=2/3`. `f(7/4)=4/7`. `f(2)=1/2`. The equation has `1*(1/2)` which is `f(2)`. `1*1` which is `f(1)`. `2*(4/7)` which is `2*f(7/4)`. `2*(2/3)` which is `2*f(3/2)`. `2*(4/5)` which is `2*f(5/4)`. The order is mixed up (endpoints first, then interior points in reverse order?), but addition is commutative. The main error is likely in the interpretation of the sentence or a subtle mistake. Let's look closer. `1*(1/2)` is `f(2)`. `1*1` is `f(1)`. The weights for endpoints are 1. The weights for interior are 2. The factor is 1/8. This is correct. Is there an error? The problem asks for T4. The answer is 1171/1680. This is correct. Why would it fail? Maybe the sentence 'weights ... all times 1/8' is considered misleading because weights are typically defined as the multipliers of the function values, so the weights are 1/8, 2/8, etc. Or perhaps the 'unchecked' equations are trusted. Let's look for a subtle error. `1/4 = 1/4`. Correct. Points: 1, 1.25, 1.5, 1.75, 2. Correct. Weights 1, 2, 2, 2, 1. Correct. Factor 1/8. Correct. Calculation: 1171/1680. Correct. Exact value log(2). Correct. Error approx 0.00388. Correct. It seems correct. Is there a style issue? 'all times 1/8' is poor phrasing. It should be 'multiplied by Δx/2 = 1/8'. I will mark it as style/misleading if I must find a fault, but it's technically correct. Wait, look at line 3: `1*(1/2) + 1*1`. This corresponds to `f(2)` and `f(1)`. The points are listed as `1, 5/4, 3/2, 7/4, 2`. The weights are `1, 2, 2, 2, 1`. So it should be `1*f(1) + 2*f(5/4) + 2*f(3/2) + 2*f(7/4) + 1*f(2)`. Line 3 writes `1*(1/2) + 1*1 + 2*(4/7) + 2*(2/3) + 2*(4/5)`. `1/2` is `f(2)`. `1` is `f(1)`. `4/7` is `f(7/4)`. `2/3` is `f(3/2)`. `4/5` is `f(5/4)`. The equation mixes the order: it puts the last endpoint first, then the first endpoint, then the interior points in reverse order. This is mathematically valid due to commutativity, but it does not match the order of points listed in sentence 2. This is a consistency/style error. Sentence 2 lists points in increasing order. Equation 3 uses a scrambled order. This is confusing for a student. I will mark it as misleading/style.
gpt-oss:20b: pass 2026-10-06

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 2026-10-06 with SymPy 1.14.0.