Riemann sums: left, right and midpoint
Problem 4.1011 · medium
Compute the right Riemann sum \( \displaystyle R_{6} \) for \( \displaystyle f(x) = 12 - x^{2} \) on \( \displaystyle [1, 4] \).
- \[ 1 \cdot \frac{1}{2} = \frac{1}{2} \]Δx = (b − a)/n.✓ Proved
- The right endpoints are x = 3/2, 2, 5/2, 3, 7/2, 4.
- \[ \frac{-4 - 1 \cdot \frac{1}{4} + 3 + \frac{23}{4} + 8 + \frac{39}{4}}{2} = \frac{89}{8} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is 15 ≈ 15.0000.
Answer \( R_{6} = \frac{89}{8} \approx 11.1250 \)
Lines: 2 proved, 2 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.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | Not checked | — | a sentence; read, not computed |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | Not checked | — | 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: inconclusive — reviewer returned a non-object
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: inconclusive 2026-10-13 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-13qwen3.6:27b-mlx: inconclusive 2026-10-13 — reviewer returned a non-object
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.