Integrals of absolute values
Problem 4.285 · easy
Evaluate \( \displaystyle \int_{-2}^{5} \left| x^{2} - 2 x - 3 \right| dx \).
- \[ x^{2} - 2 x - 3 = \left(x - 3\right) \left(x + 1\right) \]The integrand is zero at x = -1, x = 3.✓ Proved
- Between those zeros the sign does not change; where it is negative, |p(x)| = −p(x).Reviewed
- \[ \int\limits_{-2}^{-1} \left(x^{2} - 2 x - 3\right)\, dx + \int\limits_{3}^{5} \left(x^{2} - 2 x - 3\right)\, dx + \int\limits_{-1}^{3} \left(- x^{2} + 2 x + 3\right)\, dx = \frac{71}{3} \]Integrate each piece and add.✓ Proved
Answer \( \frac{71}{3} \)
Lines: 2 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 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 of |p(x)| |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the roots, splits the integral at the points where the integrand changes sign, and applies the absolute value definition correctly. The final calculation is verified.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the roots, splits the integral at the points where the integrand changes sign, and applies the absolute value definition correctly. The final calculation is verified.gpt-oss:20b: pass 2026-10-04qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the roots, determines the sign of the quadratic in each interval, and sets up the piecewise integration. The final result is correct.gpt-oss:20b: fail (misleading) 2026-10-04 — The sentence claims the sign does not change between the zeros, which is incorrect: the polynomial is negative on (-1,3) and positive elsewhere. This misstatement could mislead a student about the sign analysis.
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/absolute_value_integral, checked 2026-10-04 with SymPy 1.14.0.