The textbook I'm using to study integral calculus usually assumes for it's proofs that the function takes on only positive values. The author says that if we divide the x-axis into intervals, and pick the point in each interval for which the value of the function at that point is a minimum at the interval, then we can approximate the area under the curve using inscribed rectangles. Specifically, the author says, this would be the lower sum. The upper sum is achieved by picking the point at each interval for which the value at that point is the maximum of the interval. I was thinking: wouldn't this be the opposite for negative functions. Because the upper sum would be achieved by taking the minimum at each interval, at the lower by taking the maximum at each interval.
If I were to prove the same theorem for negative functions as well, is this the only difference that it makes?
Or is this not the case? Because if the upper sum is always the greater numerically, and the lower sum is always the less numerically. And if "negative area" is a thing, then the lower sum would still be the one that is formed by taking the minimum at each point, and the upper by taking the maximum. So I am confused with the exact definitions right now and would enjoy guidance. Thank you in advance.
You seem to be misunderstanding what minimum and maximum are for negative values; these are not measured in terms of distance from zero, but simply in terms of the $x\le y$ relation.
These are all true: $$ \min \{-1,-2,-3\} = -3 \qquad \max \{-1,-2,-3\} = -1 $$ This is because $-3 \le x$ for every $x$ in that set, and because $-1 \ge x$ for every $x$ in that set.
We do not change the lower/upper sum definition for functions with potentially negative values as a result. Minimums, maximums, upper sums, and lower sums are all about being greater or lesser numerically.