Average value of a function
Problem 4.444 · medium
Find the average value of \( \displaystyle f(x) = \frac{6}{x^{2}} \) on \( \displaystyle [1, 4] \), and every \( \displaystyle c \) in the interval with \( \displaystyle f(c) = f_{\text{ave}} \).
- \[ \int\limits_{1}^{4} \frac{6}{x^{2}}\, dx = \frac{9}{2} \]The integral over the interval.✓ Proved
- \[ \frac{3}{2} \]Divide by the length of the interval.✓ Proved
- \[ \frac{3}{2} \]c = 2 lies in [1, 4].✓ Proved
Answer \( f_{\text{ave}} = \frac{3}{2},\ c = 2 \)
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: fail (error) — The solution omits the crucial step of verifying that f(2)=f_{ave}. It only states that c=2 lies in the interval, but does not show that f(2)=3/2, which is required to confirm the Mean Value Theorem for Integrals.qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly state the Mean Value Theorem for Integrals or set up the equation f(c) = f_ave to solve for c. It merely asserts that c=2 is in the interval without showing that f(2) actually equals the average value.
Every verdict on record (4)
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to explicitly state the Mean Value Theorem for Integrals or set up the equation f(c) = f_ave to solve for c. It merely asserts that c=2 is in the interval without showing that f(2) actually equals the average value.gpt-oss:20b: fail (error) 2026-10-05 — The solution omits the crucial step of verifying that f(2)=f_{ave}. It only states that c=2 lies in the interval, but does not show that f(2)=3/2, which is required to confirm the Mean Value Theorem for Integrals.qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to show the calculation of the average value formula (integral divided by interval length) and the derivation of c from f(c) = f_ave. It merely states the final numbers without connecting them to the problem statement.gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.