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Riemann sums for double integrals

Problem 11.519 · medium

Estimate \( \displaystyle \iint_R x + y \, dA \) 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 lower left corners.

\[ 1 \cdot \frac{1}{2} \cdot 1 \cdot \frac{1}{4} = \frac{1}{8} \]
ΔA = Δx Δy.✓ Proved
The sample points: (0, 0), (0, 1/2), (1/4, 0), (1/4, 1/2), (1/2, 0), (1/2, 1/2), (3/4, 0), (3/4, 1/2).
\[ 1 \cdot \frac{1}{4} \cdot 1 \cdot \frac{1}{2} \left(0 + 1 \cdot \frac{1}{4} + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + \frac{3}{4} + \frac{3}{4} + 1 + \frac{5}{4}\right) = \frac{5}{8} \]
ΔA times the sum of f at the sample points.✓ Proved
\[ \int\limits_{0}^{1}\int\limits_{0}^{1} \left(x + y\right)\, dx\, dy = 1 \]
For comparison, the exact value is 1.✓ Proved

Answer \( \frac{5}{8} \)

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: pass
qwen3.6:27b-mlx: inconclusive — reviewer returned a non-object

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the grid dimensions, sample points, and function values for the lower-left corner Riemann sum. The arithmetic is correct and matches the stated answer.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: pass 2026-10-17
gpt-oss:20b: pass 2026-10-17

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 2026-10-19 with SymPy 1.14.0.