Average value of a function
Problem 4.599 · medium
Find the average value of \( \displaystyle f(x) = x^{2} - 3 x - 1 \) on \( \displaystyle [-2, 2] \), and every \( \displaystyle c \) in the interval with \( \displaystyle f(c) = f_{\text{ave}} \).
- \[ \int\limits_{-2}^{2} \left(x^{2} - 3 x - 1\right)\, dx = \frac{4}{3} \]The integral over the interval.✓ Proved
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Divide by the length of the interval.✓ Proved
- \[ - \frac{11}{2} + \left(\frac{3}{2} - \frac{\sqrt{129}}{6}\right)^{2} + \frac{\sqrt{129}}{2} = \frac{1}{3} \]c = 3/2 - sqrt(129)/6 lies in [-2, 2].✓ Proved
Answer \( f_{\text{ave}} = \frac{1}{3},\ c = \frac{3}{2} - \frac{\sqrt{129}}{6} \)
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: fail (error) — The solution fails to find all values of c in the interval satisfying f(c) = f_ave. The quadratic equation yields two solutions, but only one is presented. Additionally, the third line is a verification of a single value rather than a derivation of the solutions.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-07qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution fails to find all values of c in the interval satisfying f(c) = f_ave. The quadratic equation yields two solutions, but only one is presented. Additionally, the third line is a verification of a single value rather than a derivation of the solutions.gpt-oss:20b: fail (misleading) 2026-10-07 — The solution claims that the chosen c lies in the interval by evaluating an incorrect expression; it does not actually verify that c∈[−2,2] and uses a nonsensical equation for f(c).qwen3.6:27b-mlx: fail (error) 2026-10-07 — The solution fails to find all values of c in the interval satisfying f(c) = f_ave. The quadratic equation yields two roots, but only one is listed. Additionally, the third line is a verification check rather than a derivation step, and the setup lacks the explicit calculation of the average value formula.
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 2026-10-07 with SymPy 1.14.0.