<s>[INST] 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 $\int|x|dx.$
 [/INST] 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.
</s>