Limits of Riemann sums as integrals
Problem 4.1228 · medium
Write \( \displaystyle \lim_{n\to\infty} \sum_{i=1}^{n} \frac{\frac{16 i}{n} + 4}{n} \) as a definite integral, and evaluate it.
- Δx = 4/n, and x_i = 1 + 4i/n runs across [1, 5] (right endpoints). The sum is Σ f(x_i)Δx with f(x) = x.
- \[ \int\limits_{1}^{5} x\, dx = 12 \]Evaluate the integral.✓ Proved
- \[ \lim_{n \to \infty} \sum_{i=1}^{n} \frac{\frac{16 i}{n} + 4}{n} = 12 \]The limit of the sum itself, from the summation formulas, agrees.✓ Proved
Answer \( \int_{1}^{5} x\, dx = 12 \)
Lines: 2 proved, 1 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. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the Riemann sum itself with n = 200,000, within its error bound |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: inconclusive — reviewer returned a non-object
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-19 — reviewer returned a non-objectgpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-16 — The setup incorrectly identifies the function as f(x) = x. The term inside the sum is (16i/n + 4)/n, which corresponds to f(x_i)Δx where x_i = 1 + 4i/n. Substituting x_i into f(x) = x gives (1 + 4i/n)(4/n) = 4/n + 16i/n^2, which does not match the given term (16i/n + 4)/n = 16i/n^2 + 4/n. The correct function is f(x) = 4x, since 4(1 + 4i/n) = 4 + 16i/n. Although the final numerical answer 12 is coincidentally correct because ∫₁⁵ 4x dx = 4∫₁⁵ x dx = 4(12) = 48? Wait. Let's re-evaluate. ∫₁⁵ x dx = 12. The sum limit is 12. If f(x)=4x, integral is 48. The sum limit is actually 12. Let's check the sum limit directly. Sum (16i/n + 4)/n = 1/n Sum(16i/n) + 1/n Sum(4) = 16/n^2 * n(n+1)/2 + 4 = 8(1+1/n) + 4/n -> 8. Wait. The problem says the limit is 12? Let's re-read carefully. Sum_{i=1}^n (16i/n + 4)/n. Term is 16i/n^2 + 4/n. Sum is 16/n^2 * n(n+1)/2 + 4/n * n = 8(n+1)/n + 4 = 8 + 8/n + 4 = 12 + 8/n. Limit is 12. Okay, the limit is 12. Now, does ∫₁⁵ x dx = 12? Yes. Does the sum correspond to ∫₁⁵ x dx? Δx = 4/n. x_i = 1 + 4i/n. f(x_i)Δx = f(1+4i/n) * 4/n. We want this to equal (16i/n + 4)/n = 4(4i/n + 1)/n. Note that x_i = 1 + 4i/n. So 4i/n = x_i - 1. The term is 4(x_i - 1 + 1)/n = 4x_i/n. This is not f(x_i)Δx with Δx=4/n unless f(x_i) * 4/n = 4x_i/n => f(x_i) = x_i. So f(x) = x. Let's check: f(x_i)Δx = (1 + 4i/n) * (4/n) = 4/n + 16i/n^2. The given term is (16i/n + 4)/n = 16i/n^2 + 4/n. They are identical. So f(x) = x is correct. The integral is ∫₁⁵ x dx = 12. The solution is actually correct. Why did I think it was wrong? I miscalculated the substitution in my head. 4(1+4i/n) = 4 + 16i/n. Divided by n? No, the term is (16i/n + 4)/n. This is (4 + 16i/n)/n. If f(x)=x, f(x_i)Δx = (1+4i/n)(4/n) = 4/n + 16i/n^2. The given term is 4/n + 16i/n^2. They match. So the solution is correct. I will mark it pass.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: fail (error) 2026-10-15 — The setup incorrectly identifies the function as f(x) = x. The term (16i/n + 4) simplifies to 4(4i/n + 1) = 4x_i, so f(x) should be 4x. Consequently, the integral should be \int_1^5 4x \, dx, not \int_1^5 x \, dx. Although the final numerical answer 12 is correct (since \int_1^5 4x \, dx = 24 and the student likely made a compensating error or the problem statement implies a different scaling, but wait: \int_1^5 4x dx = [2x^2]_1^5 = 2(25-1) = 48. The sum limit is 48. The stated answer 12 is wrong. Let's re-evaluate. Sum = \sum (16i/n + 4)/n = 16/n^2 \sum i + 4/n \sum 1 = 16/n^2 * n(n+1)/2 + 4 = 8(1+1/n) + 4/n -> 8. Wait. \lim_{n\to\infty} \sum_{i=1}^n \frac{16i/n + 4}{n} = \lim (\frac{16}{n^2}\frac{n(n+1)}{2} + \frac{4n}{n}) = \lim (8\frac{n+1}{n} + 4) = 8+4=12. Okay, the limit is 12. My integral calculation: \int_1^5 4x dx = 48. Why the discrepancy? Ah, \Delta x = 4/n. The sum is \sum f(x_i) \Delta x. If f(x)=4x, then \sum 4x_i \frac{4}{n} = \frac{16}{n} \sum x_i. This doesn't match the form \sum \frac{16i/n+4}{n}. Let's look at the term: \frac{16i/n + 4}{n} = \frac{4(4i/n + 1)}{n}. If x_i = 1 + 4i/n, then 4i/n = x_i - 1. So term is \frac{4(x_i)}{n}. This is f(x_i) \Delta x where \Delta x = 4/n? No, \Delta x = 4/n. So f(x_i) \Delta x = f(x_i) \frac{4}{n}. We have \frac{4 x_i}{n} = x_i \frac{4}{n}. So f(x) = x. Then \int_1^5 x dx = 12. Okay, so f(x)=x IS correct. Let's re-read carefully. Term: \frac{16i/n + 4}{n}. Factor out 4/n? \frac{4}{n} (\frac{4i}{n} + 1). Let \Delta x = 4/n. Then term is \Delta x (\frac{4i}{n} + 1). Since x_i = 1 + \frac{4i}{n}, the term in parenthesis is x_i. So sum is \sum x_i \Delta x. This converges to \int_1^5 x dx = 12. The solution says f(x)=x. This is correct. The integral is \int_1^5 x dx = 12. This is correct. Why did I think it was wrong? I thought f(x)=4x. If f(x)=4x, integral is 48. But the sum is \sum \frac{16i/n+4}{n}. If we write this as \sum f(x_i) \Delta x with \Delta x = 4/n, we need \frac{16i/n+4}{n} = f(x_i) \frac{4}{n} \implies f(x_i) = \frac{16i/n+4}{4} = \frac{4i}{n} + 1 = x_i. So f(x)=x is correct. The solution is actually correct. Wait, let me double check the limit calculation. \sum_{i=1}^n \frac{16i/n + 4}{n} = \frac{16}{n^2} \sum i + \frac{4}{n} \sum 1 = \frac{16}{n^2} \frac{n(n+1)}{2} + 4 = 8 \frac{n+1}{n} + 4 \to 8+4=12. Integral \int_1^5 x dx = [x^2/2]_1^5 = 25/2 - 1/2 = 12. Everything is correct. The solution is sound.gpt-oss:20b: pass 2026-10-15
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_limit_integral, checked 2026-10-19 with SymPy 1.14.0.