the method of improper integrals allow us to deal with such problems. how? well, the way we are familiar with. how do we usually try to evaluate something at a point we can't plug in? we use limits!
so, here's how it works. we replace the bad limit point with a variable. i think "N" is the usual one, but it can be whatever. and then we use limits to evaluate the integral at that point.
example, going back to . we would write this as
now we evaluate the integral using N in place of , then we plug in the limits and take the limit as .
now, not all improper integrals exist. we can only get an answer if we get definite answer for the limit of the integral. otherwise, we say the improper integral "diverges" or "does not exist" whichever's more appropriate.
so for you problem:
now evaluate both integrals and take the limits.
the integral is divergent here
...or maybe... nah, forget it
Okay, I have a question. I graphed it and it looks like the graph of 1/x (was kind of dumb on my part). Can I not say, then that -1 to 1 will cancel eachother out, and I am left with just the integral from 1 to 2 of ln(x)?
Similar sort of concept as we used in this problem: http://www.mathhelpforum.com/math-he...-integral.html
Darboux integration (Riemann). Because we require that the function be bounded on the interval in this approach.
However, there is a way to do this integral. Since it does not fall under this definition we define .
as this limit just does not exists. So it is divergent.