<s>[INST] Is my Riemann Sum correct? For the integral: $$\int_{-1}^{5} \left( x^{2} -4 \right) dx$$
My calculations:
$$\begin{align*}\Delta x &= \frac6n\\\\
x_i &= -1 + \frac{6i}n\\\\
f(x_i) &= 1 + \frac{36i^2}{n^2} -4\\\\
A&=72
\end{align*}$$
I'm unsure if this is correct as it is my first attempt at doing this type of problem.
 [/INST] Your $\Delta x$ and $x_i$ are fine, but you've gone astray after that: $$f(x_i)=x_i^2=\left(-1+\frac{6i}n\right)^2=1-\frac{12i}n+\frac{36i^2}{n^2}\;.$$ Your Riemann sum is then 
$$\begin{align*}
\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\frac6n&=\frac6n\sum_{i=1}^n\left(1-\frac{12i}n+\frac{36i^2}{n^2}\right)\\
&=\frac6n\left(\sum_{i=1}^n1-\frac{12}n\sum_{i=1}^ni+\frac{36}{n^2}\sum_{i=1}^ni^2\right)\\
&=\frac6n\left(n-\frac{12}n\cdot\frac{n(n+1)}2+\frac{36}{n^2}\cdot\frac{n(n+1)(2n+1}6\right)\\
&=\frac6n\left(n-6(n+1)+\frac{6(n+1)(2n+1)}n\right)\\
&=\frac6n\cdot\frac{-5n^2-6n+12n^2+18n+6}n\\
&=6\left(7+\frac{12}n+\frac6{n^2}\right)\;,
\end{align*}$$
which converges to $42$ as $n\to\infty$.
</s>