I was asked this question on a quiz I received a few days ago and I was kind of confused on what the answer would be. Here it is,
Set up and find the area between $$f(x)=x^2-x$$ and $$g(x)=x-1$$
So here is what I did to solve it,
Intersection at x=1 so,
$$\int_{-\infty}^{1} ((x^2-x)-(x-1)) dx + \int_{1}^{\infty} ((x-1)-(x^2+x))dx$$ $$\int_{-\infty}^{1} (x^2-2x+1) dx + \int_{1}^{\infty} (-x^2-1)dx$$ $$ \tfrac{x^3}{3}-x^2 +x \Big|_{-\infty}^1 + \tfrac{-x^3}{3} -x \Big|_1^\infty$$
At this point, I became very confused so I wrote this down, $$0$$
Anyway, my teacher gave me credit for the problem but I was wondering, is zero correct?
I would say that yes, $0$ is correct. These two lines only intersect in one point, and so, based on the usual meaning of "between" I would say either the area between the curves is infinite, or zero. Both seem pretty reasonable conclusions, and it just depends on what one considers inside vs outside. I think that since in most cases we are assuming that the area is not infinite (our justification for taking the inside area between $x^2$ and $x+1$), zero is the better answer.
I wouldn't say $+\infty$ was wrong though. The integral you are trying to calculate comes out to $+\infty$), and that's also a valid answer to the question. I assume the teacher made a mistake while making the problem, because it doesn't properly specify the region unambiguously.