Average value over a region
Problem 11.437 · easy
Find the average value of \( \displaystyle f(x, y) = e^{x + y} \) over \( \displaystyle R = [1, 3] \times [0, 3] \).
- The average is (1/area) ∬_R f dA.Reviewed
- \[ \int\limits_{0}^{3}\int\limits_{1}^{3} e^{x + y}\, dx\, dy = - e \left(-1 + e^{2}\right) + \left(-1 + e^{2}\right) e^{4} \]The double integral.✓ Proved
- \[ - \frac{e \left(-1 + e^{2}\right)}{6} + \frac{\left(-1 + e^{2}\right) e^{4}}{6} = \frac{e \left(- e^{3} - e^{2} + 1 + e^{5}\right)}{6} \]Divide by the area 6.✓ Proved
Answer \( \frac{e \left(- e^{3} - e^{2} + 1 + e^{5}\right)}{6} \)
✓ 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 2026-10-15gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14gpt-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.