∫Calc Practice

Centers of mass and centroids

Problem 5.208 · medium

A rod on \( \displaystyle 0 \le x \le 2 \) has density \( \displaystyle \rho(x) = x^{2} + 1 \). Find its center of mass.
  1. \[ \int\limits_{0}^{2} \left(x^{2} + 1\right)\, dx = \frac{14}{3} \]
    The mass.✓ Proved
  2. \[ \int\limits_{0}^{2} x \left(x^{2} + 1\right)\, dx = 6 \]
    The moment about x = 0.✓ Proved
  3. \[ \frac{9}{7} \]
    x̄ = moment / mass.✓ Proved
Answer \( \bar{x} = \frac{9}{7} \)

✓ Nihil obstat 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. Reviewers found nothing wrong with the explanation.

The full receipt
LineStatusChecked byDetail
1✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0numerical quadrature

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: pass — The solution correctly identifies the formulas for mass and moment, computes the integrals accurately, and derives the correct center of mass.
Every verdict on record (4)
  • qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the formulas for mass and moment, computes the integrals accurately, and derives the correct center of mass.
  • gpt-oss:20b: pass 2026-10-04
  • qwen3.6:27b-mlx: pass 2026-10-04 — The solution correctly identifies the formulas for mass and moment, computes the integrals accurately, and derives the correct center of mass.
  • gpt-oss:20b: pass 2026-10-04

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/center_of_mass, checked 2026-10-04 with SymPy 1.14.0.