Average value over a region
Problem 11.411 · easy
Find the average value of \( \displaystyle f(x, y) = x^{2} + y^{2} \) over \( \displaystyle R = [0, 1] \times [0, 1] \).
- The average is (1/area) ∬_R f dA.Reviewed
- \[ \int\limits_{0}^{1}\int\limits_{0}^{1} \left(x^{2} + y^{2}\right)\, dx\, dy = \frac{2}{3} \]The double integral.✓ Proved
- \[ \frac{2}{1 \cdot 1 \cdot 3} = \frac{2}{3} \]Divide by the area 1.✓ Proved
Answer \( \frac{2}{3} \)
✓ 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 2026-10-17gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly identifies the formula for the average value, computes the integral and area accurately, and arrives at the correct result.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the definition of the average value of a function over a region. The integral calculation and division by the area are correct.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/average_value_2var, checked 2026-10-17 with SymPy 1.14.0.