If your integrand is $x^2-4$, then $\displaystyle f(x_i) = x_i^2 - 4 = \left( - 1 + \frac{6i}{n} \right)^2 - 4 = -3 -\frac{12i}{n} + \frac{36 i^2}{n^2}$,
where your $\displaystyle \Delta x = \frac6n$. Your integral then becomes : $ \begin{align} \int_{-1}^{5} \left(x^2 -4 \right) dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \Delta x \right) = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) \end{align} $ Hence, all we need is to evaluate $\displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right)$ and take the limit as $n \rightarrow \infty$.
$ \begin{align} \displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = 6 \times \displaystyle \sum_{i=0}^{n-1} \left( \frac{-3}n - \frac{12i}{n^2} + \frac{36i^2}{n^3} \right)\\ & = 6 \times \left(-3 - \frac{12 n(n-1)/2}{n^2} + \frac{36 n(n-1)(2n-1)/6}{n^3} \right)\\ & = 6 \times \left( -3 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \end{align} $ where we made use of the following summations. $ \sum_{i=0}^{n-1} 1 = n $ $ \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2} $ and $ \sum_{i=0}^{n-1} i^2 = \frac{n(n-1)(2n-1)}{6} $
Now taking the limit as $n \rightarrow \infty$, we get $ \begin{align} \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = \lim_{n \rightarrow \infty} 6 \times \left( -3 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \\ & = 6 \times \left( -3 - 6 + 12 \right) = 6 \times 3 = 18 \end{align} $ Hence, $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) = 18 \end{align} $
EDIT:
Note that if $f(x) = x^2$ and $\displaystyle x_i = - 1 + \frac{6i}{n}$, then $\displaystyle f(x_i) = x_i^2 = \left( - 1 + \frac{6i}{n} \right)^2 = 1 -\frac{12i}{n} + \frac{36 i^2}{n^2}$ where your $\displaystyle \Delta x = \frac6n$. Your integral then becomes : $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \Delta x \right) = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) \end{align} $ Hence, all we need is to evaluate $\displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right)$ and take the limit as $n \rightarrow \infty$.
$ \begin{align} \displaystyle \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = 6 \times \displaystyle \sum_{i=0}^{n-1} \left( \frac1n - \frac{12i}{n^2} + \frac{36i^2}{n^3} \right)\\ & = 6 \times \left(1 - \frac{12 n(n-1)/2}{n^2} + \frac{36 n(n-1)(2n-1)/6}{n^3} \right)\\ & = 6 \times \left( 1 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \end{align} $ where we made use of the following summations. $ \sum_{i=0}^{n-1} 1 = n $ $ \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2} $ and $ \sum_{i=0}^{n-1} i^2 = \frac{n(n-1)(2n-1)}{6} $
Now taking the limit as $n \rightarrow \infty$, we get $ \begin{align} \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) & = \lim_{n \rightarrow \infty} 6 \times \left( 1 - 6 \frac{n-1}{n} + 12 \frac{(n-1)(n-1/2)}{n^2} \right) \\ & = 6 \times \left( 1 - 6 + 12 \right) = 6 \times 7 = 42 \end{align} $ Hence, $ \begin{align} \int_{-1}^{5} x^2 dx = \lim_{n \rightarrow \infty} \sum_{i=0}^{n-1} \left(f(x_i) \times \frac{6}{n} \right) = 42 \end{align} $