Swapping the limits of integral should change the sign? I have an elementary question. In this video(Link), Sal explains why swapping the limits of integration changes the sign of the result; In his video, he reasons that "dx"s in the reverse direction must have opposite sign too (because (a-b)/n must become (b-a)/n), however, that's something I don't remember I saw anywhere when I learned Riemann integral, to indicate that there's a relation between "dx"s and the order(higher/lower) of integral limits.
I have no problem with the result of definite integrals becoming negative since, well..., that's what happens when your ceiling is at the bottom of your basement! :D HOWEVER, I think it makes sense only when f changes to -f... but about the order of limits, I think it mustn't be so. Series are just algebraic sums, so I can still divide from b to a to divisions of (a-b)/n and then instead, sum all of them FROM RIGHT TO LEFT... So for example, d1.f + d2.f + d3.f + d4.f + d5.f becomes d5.f + d4.f + d3.f + d2.f + d1.f which has a similar sign and also because it's an algebraic sum, they must have the same values as well(To not confuse anyone, by "di" I mean division number i, and let's say "di" with less "i", refers to a "dx" at a position more towards the left). So I know integrals are very old and I'm definitely wrong :D But where am I making a mistake? Thanks.
 A: Here is one way to see it.  An integral over a range can be split into $2$ integrals, each covering part of the range.  In particular, you get for any integral function $f(x)$ over the domain that $\int_{a}^{b}f(x)dx + \int_{b}^{c}f(x)dx = \int_{a}^{c}f(x)dx$. Now, as the comment says, this holds for $a \le b \le c$.  However, if you consider extending/defining this to also be true for cases where $c \lt b$, e.g., if $c = a$, you get
$$\int_{a}^{b}f(x)dx + \int_{b}^{a}f(x)dx = \int_{a}^{a}f(x)dx = 0 \tag{1}\label{eq1A}$$
This results in
$$\int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx \tag{2}\label{eq2A}$$
Here is a second way to try to see this. Note the formal definition of the Riemann integral involves using a partition of the interval $[a,b]$ and taking the upper/lower limits of the sums of the functions at those partitions, times the partition widths. If $b \lt a$, then the partition values $x_i$ would be decreasing instead of increasing, so the differences $x_{i+1} - x_{i}$ would be negative instead of positive. This means the result would the negative of the integral of the same function $f(x)$ when going from the smaller limit to the larger limit.
A: The point your explainer was trying to make is right. You're integrating a differential over an interval. Since this is an infinitesimal change in the function, the change can either be an increase or a decrease (when it is nontrivial), in each case having a corresponding positive or negative sign. Thus, reversing the direction of integration also changes the sign of the integral.
That is, first consider the finite differences in your interval $\Delta x.$ Then these differences are positive or negative according as you move in the direction of increasing or decreasing $x$ correspondingly. Going to the limit will not affect these signs.
