Riemann sums for double integrals
Problem 11.457 · medium
Estimate \( \displaystyle \iint_R x + y \, dA \) over \( \displaystyle R = [0, 1] \times [0, 2] \) with a Riemann sum, \( \displaystyle m = 4 \) subintervals in \( \displaystyle x \) and \( \displaystyle n = 2 \) in \( \displaystyle y \), sampling at the midpoint points.
- \[ 2 \cdot \frac{1}{2} \cdot 1 \cdot \frac{1}{4} = \frac{1}{4} \]ΔA = Δx Δy.✓ Proved
- The sample points: (1/8, 1/2), (1/8, 3/2), (3/8, 1/2), (3/8, 3/2), (5/8, 1/2), (5/8, 3/2), (7/8, 1/2), (7/8, 3/2).
- \[ 1 \cdot 1 \cdot \frac{1}{4} \left(\frac{5}{8} + \frac{7}{8} + \frac{9}{8} + \frac{11}{8} + \frac{13}{8} + \frac{15}{8} + \frac{17}{8} + \frac{19}{8}\right) = 3 \]ΔA times the sum of f at the sample points.✓ Proved
- \[ \int\limits_{0}^{2}\int\limits_{0}^{1} \left(x + y\right)\, dx\, dy = 3 \]For comparison, the exact value is 3.✓ Proved
Answer \( 3 \)
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: inconclusive — reviewer returned a non-objectqwen3.6:27b-mlx: pass — The solution correctly identifies the subinterval widths, midpoints, and area element. The summation calculation is correct, and the comparison with the exact integral is valid.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the subinterval widths, midpoints, and area element. The summation calculation is correct, and the comparison with the exact integral is valid.gpt-oss:20b: inconclusive 2026-10-18 — reviewer returned a non-objectqwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.