Average value over a region
Problem 11.511 · easy
Find the average value of \( \displaystyle f(x, y) = x^{2} y \) over \( \displaystyle R = [0, 2] \times [0, 2] \).
- The average is (1/area) ∬_R f dA.Reviewed
- \[ \int\limits_{0}^{2}\int\limits_{0}^{2} x^{2} y\, dx\, dy = \frac{16}{3} \]The double integral.✓ Proved
- \[ \frac{16}{2 \cdot 2 \cdot 3} = \frac{4}{3} \]Divide by the area 4.✓ Proved
Answer \( \frac{4}{3} \)
✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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: passqwen3.6:27b-mlx: pass
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-19gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the formula for the average value, computes the double integral accurately, and divides by the correct area of the region.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-16gpt-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.