Limits of Riemann sums as integrals
Problem 4.1225 · medium
Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{256 i^{3}}{n^{4}} \) as a definite integral, and evaluate it.
- Δx = 4/n, and x_i = 0 + 4i/n runs across [0, 4] (right endpoints). The sum is Σ f(x_i)Δx with f(x) = x**3.
- \[ \int\limits_{0}^{4} x^{3}\, dx = 64 \]Evaluate the integral.✓ Proved
- \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{256 i^{3}}{n^{4}} = 64 \]The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{0}^{4} x^{3}\, dx = 64 \)
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 2026-10-19 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: inconclusive 2026-10-16 — reviewer returned a non-objectgpt-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 2026-10-19 with SymPy 1.14.0.