Area between $2$ curves within a closed interval I am currently studying calculus on my own and need some help with finding the area between two curves on a graph. The two functions are:
$f(x)=1-x^2$
and
$f(x)=x^2-1$
When I try to calculate the area, I keep on getting funny answers like $-8, -4, 4$, and $-8$, which I know cannot be true since I have also plotted out a graph and would estimate the answer to be around $2.5$. Please explain to me step-by-step how to solve this using integration. Thanks.
 A: First of all, we make a plot.

To compute the area between them, we first need to know from where to where we need to integrate, that is, we need to calculate their intersections. So, we solve
$$1-x^2=x^2-1$$
And this is not to difficult, since it follows that $2x^2=2$ or $x^2=1$, which has two solutions $x\in\{-1,1\}$. Now we can do a little trick. Since $x^2-1=-(1-x^2)$, we know they're just mirrored over the $x$-axis, so we actually only have to calculate the area one of them encloses with the $x$-axis, and multiply that by $2$ (because we have one of such above the axis and one under). So we do:
$$\int_{-1}^1(1-x^2)dx=[x-\tfrac13x^3]_{-1}^1=(1-\tfrac131^3)-((-1)-\tfrac13(-1)^3)=\frac43$$
Now multiply by $2$ and we get the answer: $\frac{8}{3}$.

Note that this is not something you can always do. We could just have computed

$$\int_{-1}^{1}((1-x^2)-(x^2-1))dx$$
to get the same result. In general, if you wish to calculate the area between two functions $f$ and $g$, where $\alpha<\beta$ are intersections (that is, $f(\alpha)=g(\alpha)$ and $f(\beta)=g(\beta)$), and $g(x)<f(x)$ on $(\alpha,\beta)$, then the area between them is given by
$$\int_\alpha^\beta(f(x)-g(x))dx$$
A: The area between these $2$ parabolas is: $\displaystyle \int_{-1}^1 ((1-x^2)-(x^2-1)) dx = 4\displaystyle \int_{0}^1 (1-x^2)dx = 4\left(x-\dfrac{x^3}{3}\right)|_{x=0}^{x=1}= \dfrac{8}{3}$.
A: HINT:
You have  x-limits $ x_1, x_2 = \pm 1$ 
Also please note the curves are symmetrical w.r.t. x-axis. By symmetry only area under upper parabola upto x-axis is to be found, and then doubled.
A: One can argue by symmetry that the area enclosed above the X-axis and below it 
will be the same.
So, we can calculate the area bound by one parabola and the X-axis and simply multiply it by $2$ to get to the answer:
$$A=2\int_{-1}^1(1-x^2)dx=4\int_0^1(1-x^2)dx=4(x-\frac{x^3}{3})|^{1}_0=\frac{8}{3}$$
