Home›Calculus 3›Riemann sums for double integrals›Problem 11.394

Riemann sums for double integrals

Problem 11.394 · medium

Estimate \( \displaystyle \iint_R 3 x - y^{2} + 5 \, dA \) over \( \displaystyle R = [0, 4] \times [0, 1] \) with a Riemann sum, \( \displaystyle m = 4 \) subintervals in \( \displaystyle x \) and \( \displaystyle n = 2 \) in \( \displaystyle y \), sampling at the midpoint points.

\[ 1 \cdot \frac{1}{2} \cdot 4 \cdot \frac{1}{4} = \frac{1}{2} \]
ΔA = Δx Δy.✓ Proved
The sample points: (1/2, 1/4), (1/2, 3/4), (3/2, 1/4), (3/2, 3/4), (5/2, 1/4), (5/2, 3/4), (7/2, 1/4), (7/2, 3/4).
Reviewed
\[ 1 \cdot 1 \cdot \frac{1}{2} \left(\frac{95}{16} + \frac{103}{16} + \frac{143}{16} + \frac{151}{16} + \frac{191}{16} + \frac{199}{16} + \frac{239}{16} + \frac{247}{16}\right) = \frac{171}{4} \]
ΔA times the sum of f at the sample points.✓ Proved
\[ \int\limits_{0}^{1}\int\limits_{0}^{4} \left(3 x - y^{2} + 5\right)\, dx\, dy = \frac{128}{3} \]
For comparison, the exact value is 128/3.✓ Proved

Answer \( \frac{171}{4} \)

Lines: 3 proved, 1 reviewed. 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	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: pass — The solution correctly identifies the area of each subrectangle and the midpoint sample points. The arithmetic in the Riemann sum is correct, and the comparison with the exact integral is accurate.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the area of each subrectangle and the midpoint sample points. The arithmetic in the Riemann sum is correct, and the comparison with the exact integral is accurate.
gpt-oss:20b: pass 2026-10-17
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-15
qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer returned a non-object
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.