improper integrals( very basic)

Question

I don't quite understand what it really means. Do it mean that $f(c_i)$ could be infinite and $dx$ is very small, so you can't determine what the infinite$\times$very small is?

Answer 1

The point is that if the function is nice, then the difference between all the different $f(c_i)$ for all valid choices of $c_i$ are roughly the same. Therefore, for any given partition, if it is fine enough, changing $c_i$ to some other valid value makes very little change in the total sum.

However, if $f$ is unbounded, then you can choose $c_i$ that makes $f(c_i)$ as large as you want, which means that for any given partition, different choices of $c_i$ can make that sum be basically whatever you want. That is why the limit is not defined.

Answer 2

Take a specific example, say $\int_0^2\frac 1{(x-1)^2}dx$ One of the intervals will cover $1$. The definition says the sum must converge no matter how we choose the $c_i$ in that interval (and all the rest), but if we choose that $c$ to be $1$ the sum is not defined. Even if we do not choose exactly $1$, we can choose it close enough to $1$ that the sum is enormous. As the text says, that shows we have to do something more carefully. Presumably the next paragraphs will show what that is.