Riemann sums for double integrals
Problem 11.390 · medium
Estimate \( \displaystyle \iint_R 8 x y + 4 x + 2 y \, dA \) over \( \displaystyle R = [0, 1] \times [0, 1] \) with a Riemann sum, \( \displaystyle m = 4 \) subintervals in \( \displaystyle x \) and \( \displaystyle n = 2 \) in \( \displaystyle y \), sampling at the midpoint points.
- \[ 1 \cdot \frac{1}{2} \cdot 1 \cdot \frac{1}{4} = \frac{1}{8} \]ΔA = Δx Δy.✓ Proved
- The sample points: (1/8, 1/4), (1/8, 3/4), (3/8, 1/4), (3/8, 3/4), (5/8, 1/4), (5/8, 3/4), (7/8, 1/4), (7/8, 3/4).
- \[ 1 \cdot \frac{1}{4} \cdot 1 \cdot \frac{1}{2} \left(\frac{5}{4} + \frac{11}{4} + \frac{11}{4} + \frac{17}{4} + \frac{21}{4} + \frac{23}{4} + \frac{31}{4} + \frac{41}{4}\right) = 5 \]ΔA times the sum of f at the sample points.✓ Proved
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \left(8 x y + 4 x + 2 y\right)\, dx\, dy = 5 \]For comparison, the exact value is 5.✓ Proved
Answer \( 5 \)
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. How we verify
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 sum recomputed in exact fractions |
Reviewers
gpt-oss:20b: fail (error) — The Riemann sum in step 3 is computed incorrectly; the sum of f at the midpoints is 88, giving 11, not 5. The setup and sampling are correct, but the numerical evaluation is wrong.qwen3.6:27b-mlx: pass — The solution correctly identifies the subinterval widths, calculates the correct midpoint sample points, and sums the function values accurately to arrive at the correct Riemann sum.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the subinterval widths, calculates the correct midpoint sample points, and sums the function values accurately to arrive at the correct Riemann sum.gpt-oss:20b: fail (error) 2026-10-17 — The Riemann sum in step 3 is computed incorrectly; the sum of f at the midpoints is 88, giving 11, not 5. The setup and sampling are correct, but the numerical evaluation is wrong.qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the subinterval widths, calculates the correct midpoints for the specified grid, and computes the Riemann sum accurately. The result matches the exact integral.gpt-oss:20b: fail (error) 2026-10-15 — The sample‑point values of f are incorrect; the Riemann sum evaluates to 11, not 5. The solution misapplies the function at the midpoints.qwen3.6:27b-mlx: fail (error) 2026-10-14 — The sample points listed in step 2 are incorrect for a midpoint Riemann sum with m=4 and n=2. The midpoints should be at x = 1/8, 3/8, 5/8, 7/8 and y = 1/4, 3/4, but the solution lists them correctly? Wait, let me re-read carefully. Intervals for x: [0, 1/4], [1/4, 1/2], [1/2, 3/4], [3/4, 1]. Midpoints: 1/8, 3/8, 5/8, 7/8. Correct. Intervals for y: [0, 1/2], [1/2, 1]. Midpoints: 1/4, 3/4. Correct. The points listed are: (1/8, 1/4), (1/8, 3/4), (3/8, 1/4), (3/8, 3/4), (5/8, 1/4), (5/8, 3/4), (7/8, 1/4), (7/8, 3/4). These are indeed the correct midpoints. Let's check the function values in step 3. f(x,y) = 8xy + 4x + 2y. Point (1/8, 1/4): 8(1/32) + 4(1/8) + 2(1/4) = 1/4 + 1/2 + 1/2 = 1.25 = 5/4. Correct. Point (1/8, 3/4): 8(3/32) + 4(1/8) + 2(3/4) = 3/4 + 1/2 + 3/2 = 0.75 + 0.5 + 1.5 = 2.75 = 11/4. Correct. Point (3/8, 1/4): 8(3/32) + 4(3/8) + 2(1/4) = 3/4 + 3/2 + 1/2 = 0.75 + 1.5 + 0.5 = 2.75 = 11/4. Correct. Point (3/8, 3/4): 8(9/32) + 4(3/8) + 2(3/4) = 9/4 + 3/2 + 3/2 = 2.25 + 1.5 + 1.5 = 5.25 = 21/4. Wait, the solution lists 17/4 for the 4th term? Let's check the order. The solution lists: 5/4, 11/4, 11/4, 17/4, 21/4, 23/4, 31/4, 41/4. Let's calculate all 8 values: 1. (1/8, 1/4) -> 5/4 2. (1/8, 3/4) -> 11/4 3. (3/8, 1/4) -> 11/4 4. (3/8, 3/4) -> 8(3/8)(3/4) + 4(3/8) + 2(3/4) = 8(9/32) + 12/8 + 6/4 = 9/4 + 6/4 + 6/4 = 21/4. The solution has 17/4 here. This is an error. 5. (5/8, 1/4) -> 8(5/8)(1/4) + 4(5/8) + 2(1/4) = 5/4 + 5/2 + 1/2 = 5/4 + 10/4 + 2/4 = 17/4. The solution has 21/4 here. It seems the values for (3/8, 3/4) and (5/8, 1/4) are swapped or calculated incorrectly. 6. (5/8, 3/4) -> 8(5/8)(3/4) + 4(5/8) + 2(3/4) = 15/4 + 5/2 + 3/2 = 15/4 + 10/4 + 6/4 = 31/4. The solution has 23/4 here. Error. 7. (7/8, 1/4) -> 8(7/8)(1/4) + 4(7/8) + 2(1/4) = 7/4 + 7/2 + 1/2 = 7/4 + 14/4 + 2/4 = 23/4. The solution has 31/4 here. Error. 8. (7/8, 3/4) -> 8(7/8)(3/4) + 4(7/8) + 2(3/4) = 21/4 + 7/2 + 3/2 = 21/4 + 14/4 + 6/4 = 41/4. The solution has 41/4 here. Correct. The sum in the solution is: 5+11+11+17+21+23+31+41 = 160. 160/4 = 40. 40 * 1/8 = 5. The correct sum is: 5+11+11+21+17+31+23+41 = 160. Wait, 5+11=16. 16+11=27. 27+21=48. 48+17=65. 65+31=96. 96+23=119. 119+41=160. The set of values {5/4, 11/4, 11/4, 21/4, 17/4, 31/4, 23/4, 41/4} sums to 160/4 = 40. The solution's set of values {5/4, 11/4, 11/4, 17/4, 21/4, 23/4, 31/4, 41/4} also sums to 160/4 = 40. So the final answer is correct, but the intermediate calculation of the function values at the specific sample points is wrong (the values are permuted/miscalculated). Specifically, f(3/8, 3/4) is 21/4, not 17/4. f(5/8, 1/4) is 17/4, not 21/4. f(5/8, 3/4) is 31/4, not 23/4. f(7/8, 1/4) is 23/4, not 31/4. This is an error in the arithmetic/evaluation step, even though the final sum coincidentally matches because the set of values is just permuted? No, the set of values IS the same, just assigned to the wrong points in the list. Since the sum is commutative, the final result is correct. However, stating that f(3/8, 3/4) = 17/4 is factually false. A student might think they calculated it correctly. This is an error in the solution steps.gpt-oss:20b: pass 2026-10-14
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/double_riemann, checked 2026-10-17 with SymPy 1.14.0.