∫Calc Practice

Limits of Riemann sums as integrals

Problem 4.1053 · medium

Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{\left(1 + \frac{i - 1}{n}\right)^{2}}{n} \) as a definite integral, and evaluate it.
  1. Δx = 1/n, and x_i = 1 + 1(i − 1)/n runs across [1, 2] (left endpoints). The sum is Σ f(x_i)Δx with f(x) = x**2.
    Reviewed
  2. \[ \int\limits_{1}^{2} x^{2}\, dx = \frac{7}{3} \]
    Evaluate the integral.✓ Proved
  3. \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{\left(1 + \frac{i - 1}{n}\right)^{2}}{n} = \frac{7}{3} \]
    The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{1}^{2} x^{2}\, dx = \frac{7}{3} \)

✓ Nihil obstat Lines: 2 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa 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: pass — The solution correctly identifies the Riemann sum as corresponding to the integral of x^2 from 1 to 2 using left endpoints, and the evaluation is correct.
Every verdict on record (6)
  • qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly identifies the Riemann sum as corresponding to the integral of x^2 from 1 to 2 using left endpoints, and the evaluation is correct.
  • gpt-oss:20b: pass 2026-10-17
  • qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the Riemann sum structure, the function, and the interval of integration. The evaluation is correct.
  • gpt-oss:20b: pass 2026-10-14
  • qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly identifies the Riemann sum as corresponding to the integral of x^2 from 1 to 2 using left endpoints, and the evaluation is correct.
  • 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.