Riemann sums: left, right and midpoint
Problem 4.402 · medium
Compute the midpoint Riemann sum \( \displaystyle M_{6} \) 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 midpoint points are x = 7/6, 3/2, 11/6, 13/6, 5/2, 17/6.Reviewed
- \[ \frac{\frac{85}{36} + \frac{13}{4} + \frac{157}{36} + \frac{205}{36} + \frac{29}{4} + \frac{325}{36}}{3} = \frac{575}{54} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is 32/3 ≈ 10.6667.Reviewed
Answer \( M_{6} = \frac{575}{54} \approx 10.6481 \)
✓ 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)
qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: pass 2026-10-05
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 2026-10-05 with SymPy 1.14.0.