1. **State the problem:** We are given the limit expression $$\lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^n \left(3 \left(1 + \frac{2i}{n}\right) - 6\right)$$ and asked to express it as a definite integral and find the function $f(x)$. 2. **Recall the theorem:** The definite integral from $a$ to $b$ of $f(x)$ is given by $$\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x$$ where $$\Delta x = \frac{b - a}{n} \quad \text{and} \quad x_i = a + i \Delta x$$ 3. **Identify $\Delta x$ and $x_i$ from the limit:** Here, the factor outside the sum is $\frac{2}{n}$, so $$\Delta x = \frac{2}{n}$$ which implies $$b - a = 2$$ Inside the sum, the term is $$3 \left(1 + \frac{2i}{n}\right) - 6 = 3 + \frac{6i}{n} - 6 = \frac{6i}{n} - 3$$ 4. **Express $x_i$ in terms of $i$ and $\Delta x$:** Since $$x_i = a + i \Delta x = a + i \frac{2}{n}$$ we see that $$1 + \frac{2i}{n} = 1 + i \frac{2}{n} = x_i$$ if we set $$a = 1, \quad b = 3$$ 5. **Rewrite the function inside the sum as $f(x_i)$:** $$f(x_i) = 3x_i - 6$$ 6. **Write the definite integral:** Using the theorem, $$\lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x = \int_1^3 (3x - 6) \, dx$$ 7. **Summary:** - $a = 1$ - $b = 3$ - $f(x) = 3x - 6$ **Final answer:** $$\lim_{n \to \infty} \frac{2}{n} \sum_{i=1}^n \left(3 \left(1 + \frac{2i}{n}\right) - 6\right) = \int_1^3 (3x - 6) \, dx$$