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Average value of a function

Problem 4.447 · medium

Find the average value of $ \displaystyle f(x) = - x^{2} + x - 3 $ on $ \displaystyle [-2, 2] $, and every $ \displaystyle c $ in the interval with $ \displaystyle f(c) = f_{\text{ave}} $.

\[ \int\limits_{-2}^{2} \left(- x^{2} + x - 3\right)\, dx = - \frac{52}{3} \]
The integral over the interval.✓ Proved
\[ - \frac{13}{3} \]
Divide by the length of the interval.✓ Proved
\[ - \frac{5}{2} - \frac{\sqrt{57}}{6} - \left(\frac{1}{2} - \frac{\sqrt{57}}{6}\right)^{2} = - \frac{13}{3} \]
c = 1/2 - sqrt(57)/6 lies in [-2, 2].✓ Proved
\[ - \left(\frac{1}{2} + \frac{\sqrt{57}}{6}\right)^{2} - \frac{5}{2} + \frac{\sqrt{57}}{6} = - \frac{13}{3} \]
c = 1/2 + sqrt(57)/6 lies in [-2, 2].✓ Proved

Answer $ f_{\text{ave}} = - \frac{13}{3},\ c = \frac{1}{2} - \frac{\sqrt{57}}{6},\ \frac{1}{2} + \frac{\sqrt{57}}{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
4	✓ 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: inconclusive — reviewer returned a non-object
qwen3.6:27b-mlx: fail (error) — The solution fails to explicitly calculate the average value using the formula $f_{ave} = \frac{1}{b-a} \int_a^b f(x) dx$. Line 2 shows a tautology ($-13/3 = -13/3$) without showing the division by the interval length (4), which is the crucial step to derive the average value from the integral. Additionally, the comments on lines 3 and 4 claim to verify that $c$ lies in the interval, but the equations shown are actually verifying that $f(c) = f_{ave}$, not checking the bounds.

Every verdict on record (4)

qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to explicitly calculate the average value using the formula $f_{ave} = \frac{1}{b-a} \int_a^b f(x) dx$. Line 2 shows a tautology ($-13/3 = -13/3$) without showing the division by the interval length (4), which is the crucial step to derive the average value from the integral. Additionally, the comments on lines 3 and 4 claim to verify that $c$ lies in the interval, but the equations shown are actually verifying that $f(c) = f_{ave}$, not checking the bounds.
gpt-oss:20b: inconclusive 2026-10-05 — reviewer returned a non-object
qwen3.6:27b-mlx: fail (error) 2026-10-05 — The solution fails to explicitly calculate the average value using the formula f_ave = (1/(b-a)) * integral, instead jumping to a tautology in step 2. Furthermore, steps 3 and 4 are labeled as checking if c lies in the interval, but the equations shown are actually verifying that f(c) equals the average value, which is a logical mismatch between the text description and the mathematical content.
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.