Using the formula: $$lim_{n\to∞}\sum_{i=0}^n f(xi)(\Delta x) $$
where $$f(x)=\int_0^5(x^2 - 2)$$ $$ \Delta x = {b-a\over n} $$ $$xi = a - i \Delta x $$
$\Delta x = {5\over n} $ and $xi = {5i\over n}$
I simplify to $$lim_{n\to∞}\sum_{i=0}^n{125i^2 - 10n^2\over n^3}$$
but I'm not sure where to proceed from here, as $i$ cannot be factored out of the expression.
First, let me start by saying that you should always use $i=1$ as your first index under the sum, not $i=0$. $$\displaystyle \lim_{n\to \infty}\sum_{i=1}^n{125i^2 - 10n^2\over n^3}\\ =\lim_{n\to \infty} \sum_{i=1}^n\bigg(\frac{125i^2}{n^3}-\frac{10n^2}{n^3}\bigg) \\ =\lim_{n\to \infty} \sum_{i=1}^n\frac{125i^2}{n^3}-\sum_{i=1}^n\frac{10}{n} \\ =\lim_{n\to \infty} \frac{125}{n^3}\sum_{i=1}^ni^2-\frac{10}{n}*n \\ =\lim_{n\to \infty} \frac{125}{n^3}\bigg(\frac{n(n+1)(2n+1)}{6}\bigg) -10\\ =\frac{250}{6}-10\\ =\frac{190}{6}\\ =\frac{95}{3}$$
We can confirm that this is the correct answer by evaluating the integral directly. Indeed, $\int_0^5(x^2-2)dx = [x^3/3-2x]_0^5 = 125/3-10 = \frac{95}{3}$