Centers of mass and centroids
Problem 5.484 · medium
A rod on \( \displaystyle 0 \le x \le 5 \) has density \( \displaystyle \rho(x) = 3 x + 2 \). Find its center of mass.
- \[ \int\limits_{0}^{5} \left(3 x + 2\right)\, dx = \frac{95}{2} \]The mass.✓ Proved
- \[ \int\limits_{0}^{5} x \left(3 x + 2\right)\, dx = 150 \]The moment about x = 0.✓ Proved
- \[ \frac{60}{19} \]x̄ = moment / mass.✓ Proved
Answer \( \bar{x} = \frac{60}{19} \)
✓ 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
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 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 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly applies the definition of the center of mass for a one-dimensional object, with correct bounds and integrands.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly applies the definition of the center of mass for a one-dimensional object, with correct bounds and integrands.gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: pass 2026-10-11 — The solution correctly identifies the formulas for mass and moment, computes the integrals accurately, and derives the correct center of mass.
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 2026-10-11 with SymPy 1.14.0.