1. **Problem:** Express the limit of the sum $$S_n = \lim_{n \to \infty} \sum_{i=1}^n \frac{3i + 2i}{n^r}$$ as an integral. 2. **Formula and Explanation:** The sum $$\sum_{i=1}^n f\left(\frac{i}{n}\right) \frac{1}{n}$$ approximates the definite integral $$\int_0^1 f(x) \, dx$$ as $$n \to \infty$$. 3. **Rewrite the sum:** $$S_n = \sum_{i=1}^n \frac{3i + 2i}{n^r} = \sum_{i=1}^n \frac{5i}{n^r}$$ 4. **Express in terms of $$\frac{i}{n}$$:** $$S_n = \sum_{i=1}^n 5 \cdot \frac{i}{n} \cdot \frac{1}{n^{r-1}}$$ 5. **Consider the case when $$r=1$$:** Then, $$S_n = \sum_{i=1}^n 5 \cdot \frac{i}{n} \cdot \frac{1}{n^{0}} = \sum_{i=1}^n 5 \cdot \frac{i}{n} \cdot 1 = 5 \sum_{i=1}^n \frac{i}{n} \cdot \frac{1}{n}$$ 6. **Recognize the Riemann sum:** $$S_n = 5 \sum_{i=1}^n \left(\frac{i}{n}\right) \frac{1}{n} \to 5 \int_0^1 x \, dx$$ as $$n \to \infty$$. 7. **Evaluate the integral:** $$5 \int_0^1 x \, dx = 5 \left[ \frac{x^2}{2} \right]_0^1 = 5 \cdot \frac{1}{2} = \frac{5}{2}$$ **Final answer:** $$\lim_{n \to \infty} S_n = \frac{5}{2}$$ Note: If $$r \neq 1$$, the sum does not correspond to a standard Riemann sum over [0,1].