Average value of a function
Problem 4.298 · medium
Find the average value of \( \displaystyle f(x) = - 2 x^{2} - 2 x - 3 \) on \( \displaystyle [0, 1] \), and every \( \displaystyle c \) in the interval with \( \displaystyle f(c) = f_{\text{ave}} \).
- \[ \int\limits_{0}^{1} \left(- 2 x^{2} - 2 x - 3\right)\, dx = - \frac{14}{3} \]The integral over the interval.✓ Proved
- \[ - \frac{14}{3} \]Divide by the length of the interval.✓ Proved
- \[ - \frac{\sqrt{39}}{3} - 2 - 2 \left(- \frac{1}{2} + \frac{\sqrt{39}}{6}\right)^{2} = - \frac{14}{3} \]c = -1/2 + sqrt(39)/6 lies in [0, 1].✓ Proved
Answer \( f_{\text{ave}} = - \frac{14}{3},\ c = - \frac{1}{2} + \frac{\sqrt{39}}{6} \)
Every line of this solution was proved by the computer algebra system SymPy. 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 | numerical quadrature for the average, and f evaluated at each c |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (4)
qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: inconclusive 2026-10-04 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-04
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, checked 2026-10-04 with SymPy 1.14.0.