Integrals of absolute values
Problem 4.1022 · easy
Evaluate \( \displaystyle \int_{-4}^{2} \left| x^{2} + 3 x + 2 \right| dx \).
- \[ x^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right) \]The integrand is zero at x = -2, x = -1.✓ Proved
- Between those zeros the sign does not change; where it is negative, |p(x)| = −p(x).
- \[ \int\limits_{-2}^{-1} \left(- x^{2} - 3 x - 2\right)\, dx + \int\limits_{-4}^{-2} \left(x^{2} + 3 x + 2\right)\, dx + \int\limits_{-1}^{2} \left(x^{2} + 3 x + 2\right)\, dx = \frac{55}{3} \]Integrate each piece and add.✓ Proved
Answer \( \frac{55}{3} \)
Lines: 2 proved, 1 not checked. 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 | Not checked | — | 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: fail (misleading) — The sentence claims the sign does not change between the zeros, which is false; the quadratic is negative only on (-2,-1).qwen3.6:27b-mlx: pass — The solution correctly identifies the roots of the quadratic, determines the sign of the polynomial in each sub-interval, and sets up the piecewise integration with correct bounds and signs. The final calculation is correct.
Every verdict on record (4)
gpt-oss:20b: fail (misleading) 2026-10-13 — The sentence claims the sign does not change between the zeros, which is false; the quadratic is negative only on (-2,-1).qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the roots of the quadratic, determines the sign of the polynomial in each sub-interval, and sets up the piecewise integration with correct bounds and signs. The final calculation is correct.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the roots of the quadratic, determines the sign of the polynomial in each sub-interval, and sets up the piecewise integration correctly. The final result is accurate.
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 2026-10-13 with SymPy 1.14.0.