I understand that when evaluating
$ \int_{-1}^{2} \frac{1}{x} \mathrm dx = \ln 2$
It's simple integration, I understand. I'm more focused on the theory behind if it even exists. I had a question from Larson, Edward's Calculus 9th edition that was a true or false relating to this earlier today.
In one sense, I thought it had to be $\ln 2$. But, when looking at it again, technically the area from $-1$ to $0$, and $0$ to $1$ are negative infinity and infinity, respectively. They should cancel out and our original integral from $-1$ to $2$ is the same as the integral from $1$ to $2$. However, technically the integral does not converge from those two endpoints. The areas from $-1$ to $0$ and $0$ to $1$ are only infinitesimally close as they both approach zero from one side.
I was looking for your guys' input here. I was debating myself most of today over this seamlessly simple true/false question.
Note: this is a Calc 1 class, but I suppose I'm just getting ideas from Calc 2 onward (I've self-studied a bit) intertwined with our knowledge we've learned so far.