Average value of a function
Problem 4.294 · medium
Find the average value of \( \displaystyle f(x) = 3 x^{2} + 2 x + 3 \) on \( \displaystyle [-1, 3] \), and every \( \displaystyle c \) in the interval with \( \displaystyle f(c) = f_{\text{ave}} \).
- \[ \int\limits_{-1}^{3} \left(3 x^{2} + 2 x + 3\right)\, dx = 48 \]The integral over the interval.✓ Proved
- \[ 12 \]Divide by the length of the interval.✓ Proved
- \[ \frac{7}{3} + \frac{4 \sqrt{7}}{3} + 3 \left(- \frac{1}{3} + \frac{2 \sqrt{7}}{3}\right)^{2} = 12 \]c = -1/3 + 2*sqrt(7)/3 lies in [-1, 3].✓ Proved
Answer \( f_{\text{ave}} = 12,\ c = - \frac{1}{3} + \frac{2 \sqrt{7}}{3} \)
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.