Question about this double integral

[DottZakapa]

TL;DR

double integral

could please some one explain the inequality on the right?
in particular how should i see

and

thanks

 

[Infrared]

You are working with $0\leq x\leq 2$. If $0\leq x\leq 1$, then $\text{min}\{x^2,x\}=x^2$ and if $1\leq x\leq 2$, then $\text{min}\{x^2,x\}=x$ (and other other way around for $\text{max}$).

 

[DottZakapa]

why?

 

[Mark44]

why?

Look at the graphs of $y = x$ and $y = x^2$ on the interval [0, 2]. When $x \in [0, 1]$, which graph is higher? Same question when $x \in [1, 2]$.

 

[jedishrfu]

The $min\{x^2,x\}$ with x between 0 and 2 inclusively means comparing $x^2$ to $x$ and selecting the lesser number of the two.

As an example, if $x=-2$ then the comparison would be 4 vs -2 and so -2 is the lesser one.

In this case, $x=0.5$ vs $x^2=0.25$ so that $x^2$ is the lesser when $x$ is between 0 and 1.
and for the case, $x=1.5$ vs $x^2= 2.25$ then $x$ is the lesser one when $x$ is between 1 and 2.

 

[DottZakapa]

Aw ok now is clear, thanks a lot