∫Calc Practice
Home›Calculus 2›Trapezoidal, midpoint and Simpson's rules›Problem 4.978

Trapezoidal, midpoint and Simpson's rules

Problem 4.978 · medium

Approximate \( \displaystyle \int_{0}^{2} \frac{x}{x + 1}\, dx \) using the trapezoidal rule with \( \displaystyle n = 6 \).
  1. \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]
    Δx = (b − a)/n.✓ Proved
  2. 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.
  3. \[ \frac{1 \cdot 0 + 2 \cdot 1 \cdot \frac{1}{4} + 1 \cdot 2 \cdot \frac{1}{3} + 2 \cdot 2 \cdot \frac{1}{5} + 2 \cdot 1 \cdot \frac{1}{2} + 2 \cdot 4 \cdot \frac{1}{7} + 2 \cdot 5 \cdot \frac{1}{8}}{6} = \frac{2251}{2520} \]
    Weighted sum of the function values.✓ Proved
  4. \[ \int\limits_{0}^{2} \frac{x}{x + 1}\, dx = 2 - \ln{\left(3 \right)} \]
    The exact value, ≈ 0.901388, for comparison: the error is about 8.13e-03.✓ Proved
Answer \( T_{6} = \frac{2251}{2520} \approx 0.893254 \)

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
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the rule implemented separately in floating point

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: fail (error) — The solution incorrectly states that the trapezoidal rule weights are multiplied by 1/6. The correct factor is Δx/2, which is (1/3)/2 = 1/6 only if the weights 1, 2, ..., 1 are applied to the sum directly without the 1/2 factor in the standard formula T_n = (Δx/2)[f(x_0) + 2f(x_1) + ... + f(x_n)]. However, line 2 says 'all times 1/6', implying the formula used is (1/6) * Sum(weights * f(x)). The standard formula is (Δx/2) * Sum(...). Since Δx = 1/3, Δx/2 = 1/6. So the multiplier is indeed 1/6. Wait, let me re-read carefully. Line 2 says 'weights 1, 2, 2, 2, 2, 2, 1, all times 1/6'. This describes the calculation in Line 3: (Sum of weighted f(x)) / 6. This is correct because T_n = (Δx/2) * [f_0 + 2f_1 + ... + f_n]. Here Δx/2 = 1/6. So the arithmetic is correct. Let me check the function values in Line 3. f(x) = x/(x+1). x=0 -> 0. x=1/3 -> (1/3)/(4/3) = 1/4. x=2/3 -> (2/3)/(5/3) = 2/5. x=1 -> 1/2. x=4/3 -> (4/3)/(7/3) = 4/7. x=5/3 -> (5/3)/(8/3) = 5/8. x=2 -> 2/3. Line 3 has: 1*0 + 2*(1/4) + 1*(2/3) ... WAIT. The weight for x=2/3 (the second interior point? No, x_2) should be 2. The points are x_0=0, x_1=1/3, x_2=2/3, x_3=1, x_4=4/3, x_5=5/3, x_6=2. Weights should be 1, 2, 2, 2, 2, 2, 1. Line 3 has: 1*0 (correct for x_0) + 2*(1/4) (correct for x_1) + 1*(2/3) (INCORRECT weight for x_2, should be 2) + 2*(2/5) (correct for x_3) + 2*(1/2) (correct for x_4) + 2*(4/7) (correct for x_5) + 2*(5/8) (INCORRECT weight for x_6, should be 1). The last term is 2*(5/8) but x_6=2, f(2)=2/3. The term for x_6 is missing or mislabeled. Let's look at the terms in Line 3 again: 1*0 (x_0), 2*(1/4) (x_1), 1*(2/3) (this looks like f(2)=2/3 but with weight 1? No, 2/3 is f(2/3)? No f(2/3)=2/5. f(2)=2/3. So 1*(2/3) is likely the last term f(x_6) with weight 1. But it is placed in the middle. And 2*(5/8) is f(5/3) with weight 2. So the terms are: f(0), 2f(1/3), f(2) [wrong position/label?], 2f(1), 2f(4/3), 2f(5/3). Where is 2f(2/3)? It's missing. The term 1*(2/3) is actually f(2) with weight 1. The term for x_2=2/3 (value 2/5) is missing from the sum, or rather, the term 2*(2/5) is present. Let's list the terms in Line 3 explicitly: 1*0, 2*(1/4), 1*(2/3), 2*(2/5), 2*(1/2), 2*(4/7), 2*(5/8). Values: 0, 1/2, 2/3, 4/5, 1, 8/7, 5/4. Sum = 0 + 0.5 + 0.666 + 0.8 + 1 + 1.142 + 1.25 = 5.358. Divided by 6 = 0.893. Correct sum should be: 1*0 + 2*(1/4) + 2*(2/5) + 2*(1/2) + 2*(4/7) + 2*(5/8) + 1*(2/3). The solution has 1*(2/3) instead of 2*(2/5) for the x_2 term? No, it has 2*(2/5). It has 1*(2/3) which is f(2). It has 2*(5/8) which is 2*f(5/3). It is missing the final weight 1 for f(2) if 1*(2/3) is meant to be f(2/3)? No f(2/3)=2/5. 1*(2/3) is f(2). So the term for x_2=2/3 is 2*(2/5). The term for x_6=2 is 1*(2/3). The solution lists 1*(2/3) as the 3rd term. This is just a permutation of terms, which is fine for addition. BUT, look at the last term: 2*(5/8). This is 2*f(5/3). The last point is x_6=2. Its weight should be 1. The solution has 2*(5/8) as the last term. It does NOT have a term for x_6=2 with weight 1? Wait, 1*(2/3) is in the list. Is that x_6? Yes. So the weights are 1, 2, 1, 2, 2, 2, 2? No. The list is: 1*f(0), 2*f(1/3), 1*f(2), 2*f(1), 2*f(4/3), 2*f(5/3). It is missing 2*f(2/3)? No, 2*(2/5) is 2*f(2/3). So the terms are: f(0), 2f(1/3), f(2), 2f(1), 2f(4/3), 2f(5/3). It is missing 2f(2/3)? No, 2*(2/5) is there. It is missing f(2)? No, 1*(2/3) is there. It is missing 2f(2/3)? No. Let's check the count. 7 terms. Points: 0, 1/3, 2/3, 1, 4/3, 5/3, 2. Weights: 1, 2, 2, 2, 2, 2, 1. Solution terms: 1*f(0), 2*f(1/3), 1*f(2) [Weight 1 for x_2? No, x_2 is 2/3. f(2/3)=2/5. The term 1*(2/3) is f(2). So the solution used weight 1 for x_2? No, it used weight 1 for the value 2/3. The value 2/3 comes from x=2. So it assigned weight 1 to x=2. But it placed it in the 3rd slot. Then it has 2*(2/5) which is 2*f(2/3). Then 2*(1/2) is 2*f(1). Then 2*(4/7) is 2*f(4/3). Then 2*(5/8) is 2*f(5/3). So the weights used are: 1 for x=0, 2 for x=1/3, 1 for x=2, 2 for x=2/3, 2 for x=1, 2 for x=4/3, 2 for x=5/3. The weight for x=2 should be 1. The weight for x=5/3 should be 2. The weight for x=2/3 should be 2. The solution has 2 for x=5/3. It has 1 for x=2. It has 2 for x=2/3. It has 2 for x=1. It has 2 for x=4/3. It has 2 for x=1/3. It has 1 for x=0. All weights are correct! The order is just scrambled. 1*f(0) + 2*f(1/3) + 1*f(2) + 2*f(2/3) + 2*f(1) + 2*f(4/3) + 2*f(5/3). This sums to the correct value. So the calculation is correct. Why did I think it was wrong? I misread the term 1*(2/3) as being for x=2/3. It is for x=2. The solution is actually correct. Let me double check the 'proved' status. The equation is marked proved. The sentence says 'weights 1, 2, 2, 2, 2, 2, 1'. This implies the standard order. The equation shows a scrambled order but the same terms. Is this misleading? It might be confusing to a student to see the terms out of order, but it is not mathematically wrong. However, look at the last term in the equation: 2*(5/8). This is 2*f(5/3). The last point is x=2. The weight for x=2 is 1. The term for x=2 is 1*(2/3). It appears 3rd in the sum. This is fine. Is there any error? Let's re-read the sentence. 'The trapezoidal rule uses the points ... with weights 1, 2, 2, 2, 2, 2, 1, all times 1/6.' This is a correct description of the method. The equation implements it. The result is correct. Verdict: pass.
Every verdict on record (4)
  • gpt-oss:20b: pass 2026-10-12
  • qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly states that the trapezoidal rule weights are multiplied by 1/6. The correct factor is Δx/2, which is (1/3)/2 = 1/6 only if the weights 1, 2, ..., 1 are applied to the sum directly without the 1/2 factor in the standard formula T_n = (Δx/2)[f(x_0) + 2f(x_1) + ... + f(x_n)]. However, line 2 says 'all times 1/6', implying the formula used is (1/6) * Sum(weights * f(x)). The standard formula is (Δx/2) * Sum(...). Since Δx = 1/3, Δx/2 = 1/6. So the multiplier is indeed 1/6. Wait, let me re-read carefully. Line 2 says 'weights 1, 2, 2, 2, 2, 2, 1, all times 1/6'. This describes the calculation in Line 3: (Sum of weighted f(x)) / 6. This is correct because T_n = (Δx/2) * [f_0 + 2f_1 + ... + f_n]. Here Δx/2 = 1/6. So the arithmetic is correct. Let me check the function values in Line 3. f(x) = x/(x+1). x=0 -> 0. x=1/3 -> (1/3)/(4/3) = 1/4. x=2/3 -> (2/3)/(5/3) = 2/5. x=1 -> 1/2. x=4/3 -> (4/3)/(7/3) = 4/7. x=5/3 -> (5/3)/(8/3) = 5/8. x=2 -> 2/3. Line 3 has: 1*0 + 2*(1/4) + 1*(2/3) ... WAIT. The weight for x=2/3 (the second interior point? No, x_2) should be 2. The points are x_0=0, x_1=1/3, x_2=2/3, x_3=1, x_4=4/3, x_5=5/3, x_6=2. Weights should be 1, 2, 2, 2, 2, 2, 1. Line 3 has: 1*0 (correct for x_0) + 2*(1/4) (correct for x_1) + 1*(2/3) (INCORRECT weight for x_2, should be 2) + 2*(2/5) (correct for x_3) + 2*(1/2) (correct for x_4) + 2*(4/7) (correct for x_5) + 2*(5/8) (INCORRECT weight for x_6, should be 1). The last term is 2*(5/8) but x_6=2, f(2)=2/3. The term for x_6 is missing or mislabeled. Let's look at the terms in Line 3 again: 1*0 (x_0), 2*(1/4) (x_1), 1*(2/3) (this looks like f(2)=2/3 but with weight 1? No, 2/3 is f(2/3)? No f(2/3)=2/5. f(2)=2/3. So 1*(2/3) is likely the last term f(x_6) with weight 1. But it is placed in the middle. And 2*(5/8) is f(5/3) with weight 2. So the terms are: f(0), 2f(1/3), f(2) [wrong position/label?], 2f(1), 2f(4/3), 2f(5/3). Where is 2f(2/3)? It's missing. The term 1*(2/3) is actually f(2) with weight 1. The term for x_2=2/3 (value 2/5) is missing from the sum, or rather, the term 2*(2/5) is present. Let's list the terms in Line 3 explicitly: 1*0, 2*(1/4), 1*(2/3), 2*(2/5), 2*(1/2), 2*(4/7), 2*(5/8). Values: 0, 1/2, 2/3, 4/5, 1, 8/7, 5/4. Sum = 0 + 0.5 + 0.666 + 0.8 + 1 + 1.142 + 1.25 = 5.358. Divided by 6 = 0.893. Correct sum should be: 1*0 + 2*(1/4) + 2*(2/5) + 2*(1/2) + 2*(4/7) + 2*(5/8) + 1*(2/3). The solution has 1*(2/3) instead of 2*(2/5) for the x_2 term? No, it has 2*(2/5). It has 1*(2/3) which is f(2). It has 2*(5/8) which is 2*f(5/3). It is missing the final weight 1 for f(2) if 1*(2/3) is meant to be f(2/3)? No f(2/3)=2/5. 1*(2/3) is f(2). So the term for x_2=2/3 is 2*(2/5). The term for x_6=2 is 1*(2/3). The solution lists 1*(2/3) as the 3rd term. This is just a permutation of terms, which is fine for addition. BUT, look at the last term: 2*(5/8). This is 2*f(5/3). The last point is x_6=2. Its weight should be 1. The solution has 2*(5/8) as the last term. It does NOT have a term for x_6=2 with weight 1? Wait, 1*(2/3) is in the list. Is that x_6? Yes. So the weights are 1, 2, 1, 2, 2, 2, 2? No. The list is: 1*f(0), 2*f(1/3), 1*f(2), 2*f(1), 2*f(4/3), 2*f(5/3). It is missing 2*f(2/3)? No, 2*(2/5) is 2*f(2/3). So the terms are: f(0), 2f(1/3), f(2), 2f(1), 2f(4/3), 2f(5/3). It is missing 2f(2/3)? No, 2*(2/5) is there. It is missing f(2)? No, 1*(2/3) is there. It is missing 2f(2/3)? No. Let's check the count. 7 terms. Points: 0, 1/3, 2/3, 1, 4/3, 5/3, 2. Weights: 1, 2, 2, 2, 2, 2, 1. Solution terms: 1*f(0), 2*f(1/3), 1*f(2) [Weight 1 for x_2? No, x_2 is 2/3. f(2/3)=2/5. The term 1*(2/3) is f(2). So the solution used weight 1 for x_2? No, it used weight 1 for the value 2/3. The value 2/3 comes from x=2. So it assigned weight 1 to x=2. But it placed it in the 3rd slot. Then it has 2*(2/5) which is 2*f(2/3). Then 2*(1/2) is 2*f(1). Then 2*(4/7) is 2*f(4/3). Then 2*(5/8) is 2*f(5/3). So the weights used are: 1 for x=0, 2 for x=1/3, 1 for x=2, 2 for x=2/3, 2 for x=1, 2 for x=4/3, 2 for x=5/3. The weight for x=2 should be 1. The weight for x=5/3 should be 2. The weight for x=2/3 should be 2. The solution has 2 for x=5/3. It has 1 for x=2. It has 2 for x=2/3. It has 2 for x=1. It has 2 for x=4/3. It has 2 for x=1/3. It has 1 for x=0. All weights are correct! The order is just scrambled. 1*f(0) + 2*f(1/3) + 1*f(2) + 2*f(2/3) + 2*f(1) + 2*f(4/3) + 2*f(5/3). This sums to the correct value. So the calculation is correct. Why did I think it was wrong? I misread the term 1*(2/3) as being for x=2/3. It is for x=2. The solution is actually correct. Let me double check the 'proved' status. The equation is marked proved. The sentence says 'weights 1, 2, 2, 2, 2, 2, 1'. This implies the standard order. The equation shows a scrambled order but the same terms. Is this misleading? It might be confusing to a student to see the terms out of order, but it is not mathematically wrong. However, look at the last term in the equation: 2*(5/8). This is 2*f(5/3). The last point is x=2. The weight for x=2 is 1. The term for x=2 is 1*(2/3). It appears 3rd in the sum. This is fine. Is there any error? Let's re-read the sentence. 'The trapezoidal rule uses the points ... with weights 1, 2, 2, 2, 2, 2, 1, all times 1/6.' This is a correct description of the method. The equation implements it. The result is correct. Verdict: pass.
  • qwen3.6:27b-mlx: fail (error) 2026-10-12 — The solution incorrectly states that the weights are multiplied by 1/6. The trapezoidal rule formula is (Δx/2) * [f(x_0) + 2f(x_1) + ... + f(x_n)]. Since Δx = 1/3, the multiplier should be 1/6, but the text says 'all times 1/6' implying the sum of weights (which is 12) is multiplied by 1/6, resulting in a factor of 2 error in the logic description, although the calculation in line 3 divides by 6. Wait, let's look closer. Line 3 calculates the sum of weighted function values and divides by 6. The standard formula is T_n = (Δx/2) * (w_0 f_0 + ... + w_n f_n). Here Δx = 1/3, so Δx/2 = 1/6. The weights w_i are 1, 2, ..., 1. So the calculation in line 3 is correct: Sum(weights * f) / 6. However, sentence 2 says 'weights 1, 2, ..., 1, all times 1/6'. This phrasing is ambiguous or incorrect. Usually, one says 'sum of weighted values multiplied by Δx/2'. Saying the weights are 'all times 1/6' suggests the weights themselves are 1/6, 2/6, etc., which is not standard terminology, or it suggests the final multiplier is 1/6 applied to the sum of weights*values. The bigger issue is likely the calculation in line 3. Let's check the values. f(0)=0. f(1/3)=1/4. f(2/3)=2/5. f(1)=1/2. f(4/3)=4/7. f(5/3)=5/8. f(2)=2/3. Sum = 0 + 2(1/4) + 2(2/5) + 2(1/2) + 2(4/7) + 2(5/8) + 2/3? No, the last term is f(2)=2/3 with weight 1. Line 3 has '1*(2/3)' at the end? No, line 3 has '1*(2/3)'? Let's re-read line 3 carefully: '(1*0 + 2*(1/4) + 1*(2/3) + 2*(2/5) + 2*(1/2) + 2*(4/7) + 2*(5/8))/6'. The term for x=2/3 is given weight 1. The term for x=1 is given weight 2. The term for x=4/3 is given weight 2. The term for x=5/3 is given weight 2. The term for x=2 is missing from the sum in line 3? Or is '1*(2/3)' the last term? The points are 0, 1/3, 2/3, 1, 4/3, 5/3, 2. There are 7 points. The sum in line 3 has 7 terms. Term 1: 1*0 (x=0). Term 2: 2*(1/4) (x=1/3). Term 3: 1*(2/3) (x=2/3). This is wrong. The weight for x=2/3 should be 2. Term 4: 2*(2/5) (x=1). Weight 2 is correct. Term 5: 2*(1/2) (x=4/3). Weight 2 is correct. Term 6: 2*(4/7) (x=5/3). Weight 2 is correct. Term 7: 2*(5/8) (x=2??). f(2) = 2/3, not 5/8. f(5/3) = 5/8. So the last term in the sum is 2*(5/8), which corresponds to x=5/3. The term for x=2 is missing entirely. The sum in line 3 calculates the trapezoidal rule for n=5? Or it just drops the last point. The weights listed in sentence 2 are 1, 2, 2, 2, 2, 2, 1 (7 weights). The calculation in line 3 uses weights 1, 2, 1, 2, 2, 2, 2. It assigns weight 1 to the 3rd point (2/3) and weight 2 to the 7th point (which it evaluates as 5/8, i.e., x=5/3). It completely omits the evaluation at x=2. This is a calculation error.
  • gpt-oss:20b: pass 2026-10-12

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.

← Problem 4.977All trapezoidal, midpoint and simpson's rules problemsProblem 4.979 →