Limits of Riemann sums as integrals
Problem 4.1178 · medium
Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{\left(i - 1\right)^{2}}{n^{3}} \) as a definite integral, and evaluate it.
- Δx = 1/n, and x_i = 0 + 1(i − 1)/n runs across [0, 1] (left endpoints). The sum is Σ f(x_i)Δx with f(x) = x**2.Reviewed
- \[ \int\limits_{0}^{1} x^{2}\, dx = \frac{1}{3} \]Evaluate the integral.✓ Proved
- \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{\left(i - 1\right)^{2}}{n^{3}} = \frac{1}{3} \]The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{0}^{1} x^{2}\, dx = \frac{1}{3} \)
Lines: 2 proved, 1 reviewed. 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
| 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 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the Riemann sum itself with n = 200,000, within its error bound |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass — The solution correctly identifies the Riemann sum structure, properly handles the left-endpoint indexing by shifting the index or recognizing the limit equivalence, and evaluates the integral correctly.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18 — The solution correctly identifies the Riemann sum structure, properly handles the left-endpoint indexing by shifting the index or recognizing the limit equivalence, and evaluates the integral correctly.gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly identifies the Riemann sum structure, defines the function and interval appropriately, and verifies the result.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.