# why does the variable bound (i.e. the bound that changes) must always be upper bound

I'm currently learning integral calculus, in particularly, integral function with the lower bound been the variable and upper bound is the constant.

Below is a question and solution from khan academy

My question is,

can't we just work with the lower bound being the variable (i.e. upper bound is a constant and lower bound is a variable that changes) instead of switching the upper and lower bound (so that variable bound is the upper bound) and then add a negative sign?

It seem to me that enforcing the lower bound to be a constant and upper bound to be a variable that changes, is somewhat artificial (unless there is some fundamental difference between the two scenario that I haven't understood yet).

One other explanation might be because the fundamental theorem of calculus is usually stated as $$\frac{d}{dx} \int_a^x f(t) \, dt = f(x),$$ so solutions for questions where "$$x$$" is the lower limit of integration simply switch the bounds in order to make the expression look more like the above form so that you can use the theorem. Of course one could equivalently state the theorem as $$\frac{d}{dx} \int_x^b f(t) \, dt = -f(x)$$ instead, but maybe the minus sign looks less pleasing here.
• And if we have only the theorem with $x$ as the lower bound, we would have the question why it always has to be on the bottom. So then we answer that by making the theorem give two rules, one for the bottom and one for the top, and we have to memorize both and remember which has the negative sign. Whereas if we just remember the "top" version we can easily apply it by swapping bounds whenever needed. – David K Oct 15 '18 at 2:03