Riemann sums: left, right and midpoint
Problem 4.543 · medium
Compute the right Riemann sum \( \displaystyle R_{5} \) for \( \displaystyle f(x) = x^{3} + 1 \) on \( \displaystyle [0, 1] \).
- \[ 1 \cdot \frac{1}{5} = \frac{1}{5} \]Δx = (b − a)/n.✓ Proved
- The right endpoints are x = 1/5, 2/5, 3/5, 4/5, 1.Reviewed
- \[ \frac{\frac{126}{125} + \frac{133}{125} + \frac{152}{125} + \frac{189}{125} + 2}{5} = \frac{34}{25} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is 5/4 ≈ 1.2500.Reviewed
Answer \( R_{5} = \frac{34}{25} \approx 1.3600 \)
✓ Nihil obstat Lines: 2 proved, 2 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | a sentence; read, not computed |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the sum recomputed in 30-digit floating point |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: pass
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06gpt-oss:20b: pass 2026-10-06qwen3.6:27b-mlx: pass 2026-10-06 — The solution correctly identifies the step size, right endpoints, and function evaluations. The arithmetic is correct, and the comparison with the exact integral is accurate.
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_sums, checked 2026-10-06 with SymPy 1.14.0.