$$\lim_{n \to \infty} \frac{3}{n}\left\{4 + \left(2 + \frac{1}{n}\right)^2 + \left(2 + \frac{2}{n}\right)^2 + \ldots + \left(3 - \frac{1}{n}\right)^2\right\}$$ is equal to

We need to evaluate:

$$ \lim_{n \to \infty} \frac{3}{n}\left\{4 + \left(2 + \frac{1}{n}\right)^2 + \left(2 + \frac{2}{n}\right)^2 + \ldots + \left(3 - \frac{1}{n}\right)^2\right\} $$

Identify the Riemann sum.

The terms inside the braces can be written as:

$$ \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2 $$

where the first term ($$r=0$$) gives $$(2+0)^2 = 4$$ and the last term ($$r=n-1$$) gives $$(2 + \frac{n-1}{n})^2 = (3 - \frac{1}{n})^2$$.

Write as a limit.

$$ \lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2 $$

This can be rewritten as:

$$ 3 \cdot \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2 $$

Recognize the Riemann sum.

$$\frac{1}{n} \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right)$$ with $$f(t) = (2+t)^2$$ is a Riemann sum for $$\int_0^1 (2+t)^2 \, dt$$.

Alternatively, with the substitution $$x = 2 + t$$:

$$ \int_0^1 (2+t)^2 \, dt = \int_2^3 x^2 \, dx $$

Evaluate the integral.

$$ \int_2^3 x^2 \, dx = \left[\frac{x^3}{3}\right]_2^3 = \frac{27}{3} - \frac{8}{3} = \frac{19}{3} $$

Multiply by 3.

$$ 3 \times \frac{19}{3} = 19 $$

The answer is Option D: $$19$$.