Comparison test for improper integrals
Problem 4.1103 · medium
Use a comparison to decide whether \( \displaystyle \int_1^\infty \frac{1}{x^{2} + 3 x} \, dx \) converges or diverges.
- x² + 3x ≥ x² for x ≥ 1, so the integrand is at most 1/x². Both are positive.Reviewed
- \[ - \frac{1}{x^{2} + 3 x} + \frac{1}{x^{2}} = \frac{3}{x^{2} \left(x + 3\right)} \]The difference, which is ≥ 0 for x ≥ 1.✓ Proved
- \[ \int\limits_{1}^{\infty} \frac{1}{x^{2}}\, dx = 1 \]The comparison integral converges.✓ Proved
- A smaller positive integrand than a convergent one converges; a larger one than a divergent one diverges. So it converges.Reviewed
Answer \( \text{converges} \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the integral over each decade [10ᵏ, 10ᵏ⁺¹] for k up to 7: its pieces vanish only if it converges |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the Direct Comparison Test. The inequality x^2 + 3x >= x^2 for x >= 1 implies 1/(x^2 + 3x) <= 1/x^2, and since the integral of 1/x^2 converges, the original integral converges.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly applies the Direct Comparison Test. The inequality x^2 + 3x >= x^2 for x >= 1 implies 1/(x^2 + 3x) <= 1/x^2, and since the integral of 1/x^2 converges, the original integral converges.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality x^2 + 3x >= x^2 for x >= 1 implies 1/(x^2 + 3x) <= 1/x^2, and since the integral of 1/x^2 converges, the original integral converges.gpt-oss:20b: pass 2026-10-14qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly applies the Direct Comparison Test. The inequality 1/(x^2+3x) <= 1/x^2 is valid for x >= 1, and since the integral of 1/x^2 converges, the original integral converges.gpt-oss:20b: pass 2026-10-14
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/improper_comparison, checked 2026-10-17 with SymPy 1.14.0.