Find the area of the region
$\displaystyle
\left \{
(x,y) : x^2 \le y \le |x|
\right \}
$
Hello,
You have to find the aera between the following functions $\displaystyle x\mapsto x^2$ and $\displaystyle x\mapsto x$ on $\displaystyle [0,1]$ (because curves meet for $\displaystyle x=1$).
By a symmetry the aera of the region is $\displaystyle \mathcal{A}=2\left(\int_{0}^{1}(x-x^2)dx\right)$
And you obtain : $\displaystyle \color{blue}\boxed{\mathcal{A}=\frac{1}{3}}$
The boundaries are given by $\displaystyle y= x^2$ and $\displaystyle y= |x|$, both of which are "even functions" and symmetric about the y-axis.
And he multiplied by 2 because it is symmetric. The area from -1 to 0 is the same as the area from 0 to 1- to find the area from -1 to 1, find the area from 0 to 1 and multiply by 2.