Question

Calcule a integral definida abaixo:
$\int\limits^1_ {-1} { \sqrt{|x| - x} } \, dx$

Answer 1

Recordemos que

$|x|=\begin{cases} -x,&\mbox{si }x\ \textless \ 0\\ x,&\mbox{si }x\geq0\\ \end{cases}$

entonces

$\displaystyle \int\limits_{-1}^1\sqrt{|x|-x}\,dx=\int\limits_{-1}^0\sqrt{|x|-x}\,dx+\int\limits_{0}^1\sqrt{|x|-x}\,dx\\ \\ \int\limits_{-1}^1\sqrt{|x|-x}\,dx=\int\limits_{-1}^0\sqrt{(-x)-x}\,dx+\int\limits_{0}^1\sqrt{x-x}\,dx\\ \\ \int\limits_{-1}^1\sqrt{|x|-x}\,dx=\int\limits_{-1}^0\sqrt{-2x}\,dx\\ \\ \int\limits_{-1}^1\sqrt{|x|-x}\,dx=\sqrt{-2}\int\limits_{-1}^0x^{1/2}\,dx$

$\displaystyle \int\limits_{-1}^1\sqrt{|x|-x}\,dx=\sqrt{-2}\,\left.\left(\frac{2}{3}x^{3/2}\right)\right|_{-1}^0\\ \\ \int\limits_{-1}^1\sqrt{|x|-x}\,dx=-\frac{2}{3}\sqrt{-2}\,\sqrt{-1}^3\\ \\ \\ \boxed{\int\limits_{-1}^1\sqrt{|x|-x}\,dx=\frac{2}{3}\sqrt{2}}$