Home›Calculus 2›Trapezoidal, midpoint and Simpson's rules›Problem 4.681

Trapezoidal, midpoint and Simpson's rules

Problem 4.681 · medium

Approximate $ \displaystyle \int_{0}^{2} x^{3} + 1\, dx $ using the trapezoidal rule with $ \displaystyle n = 6 $.

\[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]
Δx = (b − a)/n.✓ Proved
The trapezoidal rule uses the points x = 0, 1/3, 2/3, 1, 4/3, 5/3, 2 with weights 1, 2, 2, 2, 2, 2, 1, all times 1/6.
\[ \frac{1 \cdot 1 + 2 \cdot 28 \cdot \frac{1}{27} + 2 \cdot 35 \cdot \frac{1}{27} + 2 \cdot 2 + 2 \cdot 91 \cdot \frac{1}{27} + 1 \cdot 9 + 2 \cdot 152 \cdot \frac{1}{27}}{6} = \frac{55}{9} \]
Weighted sum of the function values.✓ Proved
\[ \int\limits_{0}^{2} \left(x^{3} + 1\right)\, dx = 6 \]
The exact value, ≈ 6.000000, for comparison: the error is about 1.11e-01.✓ Proved

Answer $ T_{6} = \frac{55}{9} \approx 6.111111 $

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 applies the trapezoidal rule formula by dividing the weighted sum by 6 instead of multiplying by Δx/2 (which is 1/6). The s

Every verdict on record (4)

qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution incorrectly applies the trapezoidal rule formula by dividing the weighted sum by 6 instead of multiplying by Δx/2 (which is 1/6). The s
gpt-oss:20b: pass 2026-10-08
qwen3.6:27b-mlx: fail (error) 2026-10-08 — The solution incorrectly applies the trapezoidal rule formula. It divides the weighted sum by 6 instead of multiplying by Δx/2 (which is 1/6). The correct formula is T_n = (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)]. Here Δx = 1/3, so the multiplier should be 1/6. Wait, the solution says 'all times 1/6'. Let's re-read carefully. The standard formula is $T_n = \frac{\Delta x}{2} \sum w_i f(x_i)$. Here $\Delta x = 1/3$, so $\frac{\Delta x}{2} = \frac{1}{6}$. The sentence says 'weights 1, 2, ..., 1, all times 1/6'. This implies the calculation is $\frac{1}{6} \times (1\cdot f_0 + 2\cdot f_1 + \dots)$. Let's check the equation line 3: `(1*1 + 2*(28/27) + ...)/6`. This matches the description. Is the result correct? Exact integral is 6. $T_6 = \frac{1}{6} [ f(0) + 2f(1/3) + 2f(2/3) + 2f(1) + 2f(4/3) + 2f(5/3) + f(2) ]$. $f(x) = x^3+1$. $f(0)=1$. $f(1/3) = 1/27 + 1 = 28/27$. $f(2/3) = 8/27 + 1 = 35/27$. $f(1) = 2$. $f(4/3) = 64/27 + 1 = 91/27$. $f(5/3) = 125/27 + 1 = 152/27$. $f(2) = 9$. Sum inside brackets: $1 + 2(28/27) + 2(35/27) + 2(2) + 2(91/27) + 2(152/27) + 9$. $= 10 + \frac{2}{27}(28 + 35 + 91 + 152) = 10 + \frac{2}{27}(306) = 10 + \frac{612}{27} = 10 + \frac{68}{3} = \frac{30+68}{3} = \frac{98}{3}$. $T_6 = \frac{1}{6} \times \frac{98}{3} = \frac{98}{18} = \frac{49}{9} \approx 5.444$. The solution claims the result is $55/9 \approx 6.111$. Where did 55/9 come from? If we look at the equation in line 3: `(1*1 + 2*(28/27) + 2*(35/27) + 2*2 + 2*(91/27) + 1*9 + 2*(152/27))/6`. Note the last term in the sum inside the parenthesis: `1*9` corresponds to $f(2)$, but the term before it is `2*(152/27)` which is $2f(5/3)$. The term `2*(152/27)` is correct for the second-to-last point. However, look at the weights in the sentence: '1, 2, 2, 2, 2, 2, 1'. There are 7 points. The equation has: `1*1` (w=1), `2*(...)` (w=2), `2*(...)` (w=2), `2*2` (w=2), `2*(...)` (w=2), `1*9` (w=1?? No, wait). Let's count the terms in the equation's numerator: 1. `1*1` 2. `2*(28/27)` 3. `2*(35/27)` 4. `2*2` 5. `2*(91/27)` 6. `1*9` <-- This is $f(2)$ with weight 1? 7. `2*(152/27)` <-- This is $f(5/3)$ with weight 2? The order of points is $0, 1/3, 2/3, 1, 4/3, 5/3, 2$. Indices: 0, 1, 2, 3, 4, 5, 6. Weights should be 1, 2, 2, 2, 2, 2, 1. Equation terms: 1. $f(0)$ weight 1. Correct. 2. $f(1/3)$ weight 2. Correct. 3. $f(2/3)$ weight 2. Correct. 4. $f(1)$ weight 2. Correct. 5. $f(4/3)$ weight 2. Correct. 6. $f(2)$ weight 1. **Incorrect position**. $f(2)$ is the last point. 7. $f(5/3)$ weight 2. **Incorrect position**. $f(5/3)$ is the second to last point. It seems the equation swapped the last two terms or applied the wrong weight to the last term in the list structure, but actually, looking closely at the equation: `... + 2*(91/27) + 1*9 + 2*(152/27)`. The point $x=2$ is $f(2)=9$. The point $x=5/3$ is $f(5/3)=152/27$. The standard sum is $f(0) + 2f(1/3) + 2f(2/3) + 2f(1) + 2f(4/3) + 2f(5/3) + f(2)$. The equation calculates: $f(0) + 2f(1/3) + 2f(2/3) + 2f(1) + 2f(4/3) + f(2) + 2f(5/3)$. This is mathematically equivalent to the correct sum because addition is commutative. The weights are applied to the correct function values, just listed in a slightly weird order (swapping the last two terms). So the numerator sum is indeed $98/3$. Then it divides by 6. Result $98/18 = 49/9$. The equation claims the result is $55/9$. $55/9 = 110/18$. $110/18 - 98/18 = 12/18 = 2/3$. Where does the extra $2/3$ come from? Maybe the weights were wrong? If the last term was weighted 2 instead of 1? Sum = $98/3 + f(2) = 98/3 + 9 = 98/3 + 27/3 = 125/3$. $125/3 / 6 = 125/18 \approx 6.94$. No. If the first term was weighted 2? Sum = $98/3 + 1 = 101/3$. $101/18 \approx 5.61$. No. Let's check the stated answer $55/9$. $55/9 = 110/18$. Numerator needed: $110/3 = 36.66$. Actual numerator: $98/3 = 32.66$. Difference: $4/3$. Did they use Simpson's rule? Simpson's: $\frac{\Delta x}{3} [f_0 + 4f_1 + 2f_2 + 4f_3 + 2f_4 + 4f_5 + f_6]$. $\frac{1}{9} [1 + 4(28/27) + 2(35/27) + 4(2) + 2(91/27) + 4(152/27) + 9]$. Sum inside: $10 + \frac{1}{27}(112 + 70 + 182 + 608) = 10 + \frac{972}{27} = 10 + 36 = 46$. $46/9 \approx 5.11$. No. Let's look at the equation again. `(1*1 + 2*(28/27) + 2*(35/27) + 2*2 + 2*(91/27) + 1*9 + 2*(152/27))/6` Wait, I might have misread the equation's claim. The equation says `... = 55/9`. Let's calculate the LHS value. Sum = $98/3$. LHS = $(98/3) / 6 = 98/18 = 49/9$. $49/9 \neq 55/9$. So the equation is arithmetically false. The computer algebra system marked it 'unchecked' in the prompt description? No, 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)'. Wait, the prompt says 'do not re-check the algebra of a line marked proved'. Line 3 is marked `[equation, unchecked]`. Ah, the prompt says: 'Each line of the solution 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)'. This implies if it's an equation, I should trust the algebra? But the label is `unchecked`. Usually, in these tasks, `unchecked` means I should check it, or it's a placeholder. However, the instruction says 'do not re-check the algebra of a line marked proved'. It does not say 'do not check unchecked lines'. Actually, looking at the pattern, usually `proved` means CAS verified. `unchecked` might mean the user wrote it and I need to verify. But even if I assume the equation is 'true' in the context of the solution (i.e., the solution claims this equality holds), the equality is false. $49/9 \neq 55/9$. So the solution contains a false statement. Also, the sentence in line 2 says 'all times 1/6'. If the weights are 1, 2, 2, 2, 2, 2, 1, the sum is $S$. The trapezoidal rule is $\frac{\Delta x}{2} S = \frac{1/3}{2} S = \frac{1}{6} S$. So the sentence is correct in describing the method. The error is in the calculation in line 3. $49/9 \approx 5.44$. The stated answer is $55/9 \approx 6.11$. The error is significant. Therefore, the verdict is fail. Severity: error. Notes: The calculated value in line 3 is incorrect; the sum of the weighted terms is 98/3, and dividing by 6 yields 49/9, not 55/9.
gpt-oss:20b: pass 2026-10-08

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.