1. **State the problem:** We want to find which of the given limits equals the definite integral $$\int_2^5 x^2 \, dx$$. 2. **Recall the definition of a definite integral as a limit of Riemann sums:** $$\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{k=1}^n f(x_k^*) \Delta x$$ where $$\Delta x = \frac{b-a}{n}$$ and $$x_k^*$$ is a sample point in the $$k$$th subinterval. 3. **Identify the integral bounds and function:** Here, $$a=2$$, $$b=5$$, and $$f(x) = x^2$$. 4. **Calculate $$\Delta x$$:** $$\Delta x = \frac{5-2}{n} = \frac{3}{n}$$. 5. **Check each option:** - (A) $$\lim_{n \to \infty} \sum_{k=1}^n \left(2 + \frac{k}{n}\right)^2 \frac{1}{n}$$ - Here, $$\Delta x = \frac{1}{n}$$, but the interval length is 3, so this is incorrect. - (B) $$\lim_{n \to \infty} \sum_{k=1}^n \left(2 + \frac{k}{n}\right)^2 \frac{3}{n}$$ - Here, $$\Delta x = \frac{3}{n}$$, which matches the interval length. - The sample point is $$x_k^* = 2 + \frac{k}{n}$$, but since $$\Delta x = \frac{3}{n}$$, the subintervals are of length $$\frac{3}{n}$$, so the sample points should be spaced by $$\frac{3}{n}$$, not $$\frac{1}{n}$$. - (C) $$\lim_{n \to \infty} \sum_{k=1}^n \left(2 + \frac{3k}{n}\right)^2 \frac{1}{n}$$ - Here, $$x_k^* = 2 + \frac{3k}{n}$$ and $$\Delta x = \frac{1}{n}$$. - The sample points increase by $$\frac{3}{n}$$ but the subinterval length is $$\frac{1}{n}$$, inconsistent. - (D) $$\lim_{n \to \infty} \sum_{k=1}^n \left(2 + \frac{3k}{n}\right)^2 \frac{3}{n}$$ - Here, $$\Delta x = \frac{3}{n}$$ matches the interval length. - The sample points are $$x_k^* = 2 + \frac{3k}{n}$$, which correctly partitions the interval $$[2,5]$$ into $$n$$ subintervals of length $$\frac{3}{n}$$. 6. **Conclusion:** Option (D) correctly represents the Riemann sum for $$\int_2^5 x^2 \, dx$$. 7. **Final answer:** (D) $$\boxed{\lim_{n \to \infty} \sum_{k=1}^n \left(2 + \frac{3k}{n}\right)^2 \frac{3}{n} = \int_2^5 x^2 \, dx}$$