I'm used to integrate normal functions, but here I got quiet confused because these integrals :
$\int (|1+x|-|1-x|) dx $
$\int$ max {${1-x^2,0} $}
Include absolute value, and and option to choose between two values, in what direction I must think here? do I have to separate the first integral into cases? and see whats the integral inside the absolute value is negative or positive? and in second one I just do integral on both sides normally? any kind of help would be appreciated.
On
$\int(|1+x|-|1-x|)~dx$
$=\int((1+x)\text{sgn}(1+x)-(1-x)\text{sgn}(1-x))~dx$
$=\dfrac{(1+x)^2\text{sgn}(1+x)+(1-x)^2\text{sgn}(1-x)}{2}+C$
$=\dfrac{(1+x)|1+x|+(1-x)|1-x|}{2}+C$
$\int\max\{1-x^2,0\}~dx$
$=\begin{cases}-\dfrac{2}{3}&\text{when}~x\leq-1\\x-\dfrac{x^3}{3}&\text{when}~-1\leq x\leq1\\\dfrac{2}{3}&\text{when}~x\geq1\end{cases}+C$
For the first function:
if $x\leq -1$ then $|1+x|-|1-x|=-2$
if $x\geq 1$ then $|1+x|-|1-x|=2$
if $-1 \leq x \leq 1$ then $|1+x|-|1-x|=2x$
For the second function:
if $-1\leq x \leq 1$ then $\max \{1-x^2,0\}=1-x^2$
if $x\leq -1$ or $ x \geq 1$ then $\max \{1-x^2,0\}=0$.
Using these relationships you may simply proceed with integration.