∫Calc Practice

Limits of Riemann sums as integrals

Problem 4.1054 · medium

Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{4 \left(4 i - 4\right)^{2}}{n^{3}} \) as a definite integral, and evaluate it.
  1. Δx = 4/n, and x_i = 0 + 4(i − 1)/n runs across [0, 4] (left endpoints). The sum is Σ f(x_i)Δx with f(x) = x**2.
  2. \[ \int\limits_{0}^{4} x^{2}\, dx = \frac{64}{3} \]
    Evaluate the integral.✓ Proved
  3. \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{4 \left(4 i - 4\right)^{2}}{n^{3}} = \frac{64}{3} \]
    The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{0}^{4} x^{2}\, dx = \frac{64}{3} \)

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. 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
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the Riemann sum itself with n = 200,000, within its error bound

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (6)
  • qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer returned a non-object
  • gpt-oss:20b: pass 2026-10-17
  • qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer returned a non-object
  • gpt-oss:20b: pass 2026-10-14
  • qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer returned a non-object
  • 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/riemann_limit_integral, checked 2026-10-17 with SymPy 1.14.0.