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Limits of Riemann sums as integrals

Problem 4.1223 · medium

Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{6 \left(\frac{2 i}{n} - 1\right)^{2} - 2}{n} \) as a definite integral, and evaluate it.

Δx = 2/n, and x_i = -1 + 2i/n runs across [-1, 1] (right endpoints). The sum is Σ f(x_i)Δx with f(x) = 3*x**2 - 1.
\[ \int\limits_{-1}^{1} \left(3 x^{2} - 1\right)\, dx = 0 \]
Evaluate the integral.✓ Proved
\[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{6 \left(\frac{2 i}{n} - 1\right)^{2} - 2}{n} = 0 \]
The limit of the sum itself, from the summation formulas, agrees.✓ Proved

Answer \( \int_{-1}^{1} 3 x^{2} - 1\, dx = 0 \)

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: pass
qwen3.6:27b-mlx: inconclusive — reviewer returned a non-object

Every verdict on record (6)

qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-19
qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer returned a non-object
gpt-oss:20b: pass 2026-10-16
qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer returned a non-object
gpt-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 2026-10-19 with SymPy 1.14.0.