Answer is: option1

Solution:

The general form of a Riemann sum is:

\[ \lim_{n \to \infty} \sum_{i=1}^n f\left(a + \frac{(b-a)i}{n}\right) \cdot \frac{b-a}{n} \]

The given sum is:

\[ \sum_{i=1}^n \left(-1 + \frac{3i}{n}\right)^2 \cdot \frac{3}{n} \]

1. The term \( \frac{3}{n} \) corresponds to \( \frac{b - a}{n} \), so:

\[ b - a = 3 \]

1. The term \( \left(-1 + \frac{3i}{n} \right)^2 \) corresponds to \( f\left(a + \frac{(b - a)i}{n} \right) \), so:

\[ f\left(a + \frac{3i}{n}\right) = \left(-1 + \frac{3i}{n} \right)^2 \]

Let’s choose \( a = -1 \) (a common starting point for simplicity). Then:

\[ b - (-1) = 3 \Rightarrow b = 2 \]

Now, the function argument becomes:

\[ f\left(-1 + \frac{3i}{n} \right) = \left(-1 + \frac{3i}{n} \right)^2 \]

This suggests that:

\[ f(x) = x^2 \]

Substituting \( f(x) = x^2 \), \( a = -1 \), and \( b = 2 \) into the Riemann sum, the limit becomes:

\[ \int_{-1}^2 x^2 \, dx \]

This matches option (A).