Average value over a region
Problem 11.540 · easy
Find the average value of \( \displaystyle f(x, y) = e^{x + y} \) over \( \displaystyle R = [0, 1] \times [0, 2] \).
- The average is (1/area) ∬_R f dA.Reviewed
- \[ \int\limits_{0}^{2}\int\limits_{0}^{1} e^{x + y}\, dx\, dy = - e + 1 + \left(-1 + e\right) e^{2} \]The double integral.✓ Proved
- \[ - \frac{e}{2} + \frac{1}{2} + \frac{\left(-1 + e\right) e^{2}}{2} = - \frac{e^{2}}{2} - \frac{e}{2} + \frac{1}{2} + \frac{e^{3}}{2} \]Divide by the area 2.✓ Proved
Answer \( - \frac{e^{2}}{2} - \frac{e}{2} + \frac{1}{2} + \frac{e^{3}}{2} \)
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: passqwen3.6:27b-mlx: pass — The solution correctly applies the definition of the average value of a function over a region. The setup, integration, and division by area are all correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-23 — The solution correctly applies the definition of the average value of a function over a region. The setup, integration, and division by area are all correct.gpt-oss:20b: pass 2026-10-23qwen3.6:27b-mlx: pass 2026-10-22 — The solution correctly applies the definition of the average value of a function over a region. The setup, integration, and division by area are all correct.gpt-oss:20b: pass 2026-10-22qwen3.6:27b-mlx: fail (error) 2026-10-22 — The computed double integral in line 2 is incorrect. The correct value of \(\int_0^1 \int_0^2 e^{x+y} \, dy \, dx\) is \((e-1)(e^2-1) = e^3 - e^2 + e - 1\), whereas the solution claims it equals \(e^3 - e^2 - e + 1\). This sign error in the linear terms leads to an incorrect final answer.gpt-oss:20b: pass 2026-10-22
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 2026-10-23 with SymPy 1.14.0.