∫Calc Practice

Comparison test for improper integrals

Problem 4.1102 · medium

Use a comparison to decide whether \( \displaystyle \int_1^\infty \frac{1}{x^{2} + 4 x} \, dx \) converges or diverges.
  1. x² + 4x ≥ x² for x ≥ 1, so the integrand is at most 1/x². Both are positive.
  2. \[ - \frac{1}{x^{2} + 4 x} + \frac{1}{x^{2}} = \frac{4}{x^{2} \left(x + 4\right)} \]
    The difference, which is ≥ 0 for x ≥ 1.✓ Proved
  3. \[ \int\limits_{1}^{\infty} \frac{1}{x^{2}}\, dx = 1 \]
    The comparison integral converges.✓ Proved
  4. A smaller positive integrand than a convergent one converges; a larger one than a divergent one diverges. So it converges.
Answer \( \text{converges} \)

Lines: 2 proved, 2 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. How we verify

The full receipt
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4Not checked—a sentence; read, not computed
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the integral over each decade [10ᵏ, 10ᵏ⁺¹] for k up to 7: its pieces vanish only if it converges

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: inconclusive — reviewer response could not be parsed: { "verdict": "pass", "severity": "", "notes": "The solution correctly applies the Direct Comparison Test. The inequality $x^2 + 4x \ge x^2$ for $x \ge 1$ implies $\frac{1}{x^2 + 4x} \le \frac{1}{x^2}$
Every verdict on record (6)
  • qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: { "verdict": "pass", "severity": "", "notes": "The solution correctly applies the Direct Comparison Test. The inequality $x^2 + 4x \ge x^2$ for $x \ge 1$ implies $\frac{1}{x^2 + 4x} \le \frac{1}{x^2}$
  • gpt-oss:20b: pass 2026-10-17
  • qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "pass", "severity": "", "notes": "The solution correctly applies the Direct Comparison Test. The inequality $x^2 + 4x \ge x^2$ for $x \ge 1$ implies $\frac{1}{x^2 + 4x} \le \frac{1}{x^2}$
  • gpt-oss:20b: pass 2026-10-14
  • qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "pass", "severity": "", "notes": "The solution correctly identifies the inequality $x^2 + 4x \ge x^2$ for $x \ge 1$, leading to $\frac{1}{x^2+4x} \le \frac{1}{x^2}$. It correctly applies
  • 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.