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Average value over a region

Problem 11.507 · easy

Find the average value of \( \displaystyle f(x, y) = x y \) over \( \displaystyle R = [-1, 1] \times [0, 2] \).

The average is (1/area) ∬_R f dA.
Reviewed
\[ \int\limits_{0}^{2}\int\limits_{-1}^{1} x y\, dx\, dy = 0 \]
The double integral.✓ Proved
\[ \frac{0}{2 \cdot 2} = 0 \]
Divide by the area 4.✓ Proved

Answer \( 0 \)

Lines: 2 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	Reviewed	reviewers: gpt-oss:20b, qwen3.6:27b-mlx	a sentence; read, not computed
2	✓ Proved	sympy 1.14.0	simplify(lhs - rhs) reduced to 0
3	✓ 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	two-dimensional numerical quadrature

Reviewers

gpt-oss:20b: pass
qwen3.6:27b-mlx: pass — The solution correctly applies the definition of average value. The integral is zero due to symmetry (odd function in x over symmetric interval), and the area calculation is correct.

Every verdict on record (6)

qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly applies the definition of average value. The integral is zero due to symmetry (odd function in x over symmetric interval), and the area calculation is correct.
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-16

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/average_value_2var, checked 2026-10-19 with SymPy 1.14.0.