Limits of Riemann sums as integrals
Problem 4.1453 · medium
Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{-2 + \frac{6 \left(2 i - 2\right)^{2}}{n^{2}}}{n} \) as a definite integral, and evaluate it.
- Δx = 2/n, and x_i = 0 + 2(i − 1)/n runs across [0, 2] (left endpoints). The sum is Σ f(x_i)Δx with f(x) = 3*x**2 - 1.
- \[ \int\limits_{0}^{2} \left(3 x^{2} - 1\right)\, dx = 6 \]Evaluate the integral.✓ Proved
- \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{-2 + \frac{6 \left(2 i - 2\right)^{2}}{n^{2}}}{n} = 6 \]The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{0}^{2} 3 x^{2} - 1\, dx = 6 \)
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
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | 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: inconclusive — reviewer returned a non-object
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-11-07 — reviewer returned a non-objectgpt-oss:20b: pass 2026-11-07qwen3.6:27b-mlx: inconclusive 2026-11-06 — reviewer returned a non-objectgpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: fail (error) 2026-11-06 — The setup incorrectly identifies the function as f(x) = 3x^2 - 1. The term inside the sum is -2 + 6(2(i-1)/n)^2 / n, which corresponds to f(x_i) * (1/n) where x_i = 2(i-1)/n. Since dx = 2/n, we have 1/n = dx/2. Thus, the integrand is f(x) * (1/2) = (3x^2 - 1)/2, not 3x^2 - 1. The correct integral is \int_0^2 (3x^2 - 1)/2 dx = 3, not 6.gpt-oss:20b: pass 2026-11-06
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 2026-11-07 with SymPy 1.14.0.