Riemann sums: left, right and midpoint
Problem 4.1009 · medium
Compute the right Riemann sum \( \displaystyle R_{6} \) for \( \displaystyle f(x) = \frac{1}{x} \) on \( \displaystyle [2, 4] \).
- \[ 1 \cdot \frac{1}{3} = \frac{1}{3} \]Δx = (b − a)/n.✓ Proved
- The right endpoints are x = 7/3, 8/3, 3, 10/3, 11/3, 4.Reviewed
- \[ \frac{1 \cdot \frac{1}{4} + \frac{3}{11} + \frac{3}{10} + 1 \cdot \frac{1}{3} + \frac{3}{8} + \frac{3}{7}}{3} = \frac{18107}{27720} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is -log(2) + log(4) ≈ 0.6931.Reviewed
Answer \( R_{6} = \frac{18107}{27720} \approx 0.6532 \)
✓ 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 — The solution correctly identifies the step size, the right endpoints, and the resulting sum. The comparison with the exact integral is accurate and helpful context.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the step size, the right endpoints, and the resulting sum. The comparison with the exact integral is accurate and helpful context.gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: pass 2026-10-13 — The solution correctly identifies the step size, the right endpoints, and computes the sum accurately. The comparison with the exact integral is also correct.
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 2026-10-13 with SymPy 1.14.0.