Riemann sums: left, right and midpoint
Problem 4.697 · medium
Compute the left Riemann sum \( \displaystyle L_{5} \) for \( \displaystyle f(x) = \frac{2}{x} \) on \( \displaystyle [2, 4] \).
- \[ \frac{2}{5} \]Δx = (b − a)/n.✓ Proved
- The left endpoints are x = 2, 12/5, 14/5, 16/5, 18/5.
- \[ \frac{2 \left(\frac{5}{9} + \frac{5}{8} + \frac{5}{7} + \frac{5}{6} + 1\right)}{5} = \frac{1879}{1260} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is -2*log(2) + 2*log(4) ≈ 1.3863.
Answer \( L_{5} = \frac{1879}{1260} \approx 1.4913 \)
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 2026-10-08qwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer returned a non-objectqwen3.6:27b-mlx: inconclusive 2026-10-08 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-08
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 2026-10-08 with SymPy 1.14.0.