Riemann sums: left, right and midpoint
Problem 4.404 · medium
Compute the right Riemann sum \( \displaystyle R_{5} \) for \( \displaystyle f(x) = \frac{4}{x} \) on \( \displaystyle [1, 3] \).
- \[ \frac{2}{5} \]Δx = (b − a)/n.✓ Proved
- The right endpoints are x = 7/5, 9/5, 11/5, 13/5, 3.
- \[ \frac{2 \left(\frac{4}{3} + \frac{20}{13} + \frac{20}{11} + \frac{20}{9} + \frac{20}{7}\right)}{5} = \frac{176024}{45045} \]Δx times the sum of the function values there.✓ Proved
- For comparison, the exact integral is 4*log(3) ≈ 4.3944.
Answer \( R_{5} = \frac{176024}{45045} \approx 3.9077 \)
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: fail (error) — The arithmetic in step 3 is incorrect; the correct right Riemann sum is 3.89968… (≈176024/45045 is not the exact value).qwen3.6:27b-mlx: pass — The solution correctly identifies the step size, right endpoints, and function values. The final sum is computed correctly.
Every verdict on record (4)
qwen3.6:27b-mlx: pass 2026-10-05 — The solution correctly identifies the step size, right endpoints, and function values. The final sum is computed correctly.gpt-oss:20b: fail (error) 2026-10-05 — The arithmetic in step 3 is incorrect; the correct right Riemann sum is 3.89968… (≈176024/45045 is not the exact value).qwen3.6:27b-mlx: pass 2026-10-05gpt-oss:20b: fail (error) 2026-10-05 — The computed Riemann sum is incorrect; the correct value is 3.8677… not 3.9077. The arithmetic in step 3 is flawed.
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.