Area of curves that intersect at more than two points I'm trying to set up a definite integral but the examples shown in my textbook and the way that my textbook solved the problem doesn't match.  
The problem I'm trying to solve is $y_1= 3(x^3-x)$ and $y_2 = 0$.
The way that the textbook solved this problem looks like this:

$ A = 2\int_{-1}^0 3(x^3-x)dx = 6\int_{-1}^0 (x^3-x)dx$ or $-6\int_0^1(x^3-x)dx$

Where I'm confused is where did they get the 2 that is in front of $\int_{-1}^0$?  
The textbooks shows an example that solves area of $f(x)=3x^3-x^2-10x$ and $g(x)=-x^2+2x$ like so:

$A = \int_{-2}^0[f(x)-g(x)]dx + \int_0^2[g(x)-f(x)]dx$  

But the way they setup the definite integral in the example doesn't seem to be same as the problem. Could someone elaborate why is that? And if possible in plain English would great.
 A: Draw the graph of $y=f(x)=x^3-x$, or have software do it for you.  This is very important, for if you don't what is written below will be "just words."
Note that $f(-x)=-f(x)$ for all $x$. That means that the graph for $x$ negative is the graph for positive $x$, rotated through $180^\circ$, or equivalently reflected in a point mirror at the origin. The computation you quoted takes advantage of the symmetry.
From $x=-1$ to $x=0$, our function $f(x)$ is positive, the curve is above the $x$-axis. From $0$ to $1$, we have $f(x)\le 0$. So to find the total area of the region caught between the curve and the $x$-axis, we find the area of the chunk from $-1$ to $0$, and double.  
Another way of doing it is to note that between $-1$ and $0$, $f(x)$ is above the $x$-axis, and from $0$ to $1$ it is below. So the total area is
$$\int_{-1}^0 \left((x-x^3)-0\right)\,dx + \int_{0}^1 \left(0-(x-x^3)\right)\,dx.$$
The two integrals are equal, so we might as well only compute one of them and double the result. 
