Does the integral of a function exist at a sharp point in the function? This is a pretty basic and easy question to answer, but I am not certain about its answer (I am still studying at highschool).
Let's say we have a function with some sharp point and we want to find its improper integral, such as: f(x)=|1-x| (the answer will definitely imply defining the integral as a piece-wise function). Since there's a sharp point at x=1, would that value be included in the domain of the integral of the function?
The answer in my book to this exercise (and others which are almost the same) does include those values in the domain of the integral, but this doesn't make much sense, does it? On what grounds would that be the correct solution? I thought sharp points cannot possibly be derived nor integrated. I hope I have expressed my problem clearly. Thanks in advance.
 A: A fundamental difference between differentiation and integration is that differentiation is performed at specific points, while integration is done over a region. So while you are correct that a function cannot be differentiated at "sharp points" (more formally, points where the function is not differentiable), there is nothing to stop you from integrating over sharp points (so long as your function is integrable on whatever domain you're integrating over).
Though harder to actually compute, integration is, in a sense, nicer than differentiation. A differentiable function is necessarily continuous, but this is not the case for integrable functions - in fact, any continuous function is integrable. So even if you do have "sharp points", so long as your function is continuous (and even if it's not in pretty much every case you'll ever encounter in high school), you can integrate over any points at which your function isn't differentiable.
Practically, if your domain of integration includes the point $x=1$, then that just means you'll need to split up your integral. For example:
$$\int_0^2|1-x|\,dx=\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx.$$
