Find the area of ​the figure that is bound by the lines: $y=|x^2-1|$ and $y=3|x|-3$? 
Find the area of ​​the figure that is bound by the lines:
$y=|x^2-1|$ and $y=3|x|-3$ 
I tried: $$A=\int_{-1}^{1}(|x^2-1|-3|x|+3)dx$$
Is it correct? Please, help for the right solution.
I dont know, how can I evaluate $\int|x|dx.$
 A: You should draw the graphs of the two given curves carefully before solving this type of problems, and to do that you have two find the points at which they intersect each other. Let's do that

First write the equation of two given curves explicitly-
$
y=|x^2-1|=
\begin{cases}
x^2-1, & \text{if $x\in(-\infty,-1]\cup[1,\infty)$}\\
1-x^2, & \text{if $x\in(-1,1)$}
\end{cases}
$
$
y=3|x|-3=
\begin{cases}
3x-3, & \text{if $x\ge0$}\\
-3x-3, & \text{if $x<0$}
\end{cases}
$
Now, find the points at which the cut each other.
Case 1: $x\ge1$
The equation of two curves become
$y=x^2-1$ and $y=3x-3$
A simple calculation shows that these two curves intersects at the points $(1,0)$ and $(2,3)$ in $[1,\infty)$.
Case 2: $x\in[0,1)$
The equation of two curves become
$y=1-x^2$ and $y=3x-3$
These two curves don't intersect each other in $[0,1)$
Case 3: $x\in(-1,0)$
The equation of two curves become
$y=1-x^2$ and $y=-3x-3$
These two curves don't intersect each other in $(-1,0)$
Case 4: $x\le-1$
The equation of two curves become
$y=x^2-1$ and $y=-3x-3$
A simple calculation shows that these two curves intersects at the points $(-1,0)$ and $(-2,3)$ in $(-\infty,-1]$.
So, combining all the cases we get, two curves intersect each other at the points $A(-2,3)$, $B(-1,0)$, $C(1,0)$ and $D(2,3)$.
Note one thing the curves $y=|x^2-1|$, $y=3|x|-3$ intersect $Y$-axis at $E(0,1)$, $F(0,-3)$ respectively.

So, the area of the region bounded by these two curves on the plane=
Area of the region $(ABA+BCEB+\Delta BFC+CDC)=$
$
=\int_{-2}^{-1} \lbrace(-3x-3)-(x^2-1)\rbrace dx\quad+\int_{-1}^{1} (1-x^2) dx\quad+\frac{1}{2} \times 2 \times 3\quad+\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx
$
$
=2\int_{1}^{2} \lbrace(3x-3)-(x^2-1)\rbrace dx\quad+2\int_{0}^{1} (1-x^2) dx\quad+3
$
$
=\frac{1}{3}+\frac{4}{3}+3=\frac{14}{3}
$
So, the area of the region bounded by the curves is $\frac{14}{3}$ units.
A: Note that the area bounded by the equations is between $x=-1$ and $x=1$ and that $\left|x^2-1\right|=1-x^2$ on the interval $(-1,1)$. Also note that $3\left|x\right|-3=-3x-3$ for $x<0$ and $3x-3$ for $x\geq0$. Then your integrals become:
$$\int_{-1}^0\left(1-x^2\right)-\left(-3x-3\right)\text dx+\int_0^1\left(1-x^2\right)-\left(3x-3\right)\text dx$$
$$=\int_{-1}^0-x^2+3x+4\text dx+\int_0^1-x^2-3x+4\text dx$$
which are simple to evaluate. Alternatively, as астон вілла олоф мэллбэрг mentioned, you could note that the area has even symmetry and simply double one of the integrals, which will give you the same area.
A: HINT.- By clear symmetrie you have the required area is equal to twice
$$\int_0^1(x^2-3x+2)dx+\int_1^2(3x-x^2-2)dx$$
