Riemann sums: left, right and midpoint
Problem 4.548 · medium
Compute the left Riemann sum \( \displaystyle L_{6} \) for \( \displaystyle f(x) = x^{2} + 1 \) on \( \displaystyle [1, 3] \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The left endpoints are x = 1, 4/3, 5/3, 2, 7/3, 8/3.Reviewed
- \[ \frac{2 + \frac{25}{9} + \frac{34}{9} + 5 + \frac{58}{9} + \frac{73}{9}}{3} = \frac{253}{27} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is 32/3 ≈ 10.6667.Reviewed
Answer \( L_{6} = \frac{253}{27} \approx 9.3704 \)
✓ 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
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.