Question

$\int_{-1}^{1}\sqrt{|x|-x}dx=?$

Answer 1

Resposta:

$\int_{-1}^1\sqrt{|x|-x}\;dx=\frac{2\sqrt{2}}{3}$

Explicação passo a passo:

$\int_{-1}^1\sqrt{|x|-x}\;dx=\int_{-1}^0\sqrt{|x|-x}\;dx+\int_{0}^1\sqrt{|x|-x}\;dx$

No intervalo $-1\leq x\leq 0$ temos que $|x|=-x$ e no intervalor $0\leq x\leq1$ temos que $|x|=x$, logo:

$\int_{-1}^1\sqrt{|x|-x}\;dx=\int_{-1}^0\sqrt{-x-x}\;dx+\int_{0}^1\sqrt{x-x}\;dx$

$\int_{-1}^1\sqrt{|x|-x}\;dx=\int_{-1}^0\sqrt{-2x}\;dx+\int_{0}^10\;dx$

$\int_{-1}^1\sqrt{|x|-x}\;dx=\int_{-1}^0\sqrt{-2x}\;dx$

Considerando que $u=-2x$, temos que $du/dx=-2\iff dx=(-1/2)du$. Da mesma forma, $x=-1\Rightarrow u=2$ e $x=0\Rightarrow u=0$. Temos então que:

$\int_{-1}^1\sqrt{|x|-x}\;dx=\int_{2}^0\sqrt{u}\cdot(-\frac{1}{2})\;du$

$\int_{-1}^1\sqrt{|x|-x}\;dx=-\frac{1}{2}\int_{2}^0\sqrt{u}\;du$

$\int_{-1}^1\sqrt{|x|-x}\;dx=-\frac{1}{2}\left[\frac{2}{3}x^{3/2}\right]_2^0$

$\int_{-1}^1\sqrt{|x|-x}\;dx=-\frac{1}{3}\left[x^{3/2}\right]_2^0$

$\int_{-1}^1\sqrt{|x|-x}\;dx=-\frac{1}{3}(0^{3/2}-2^{3/2})$

$\int_{-1}^1\sqrt{|x|-x}\;dx=\frac{2^{3/2}}{3}=\frac{2\sqrt{2}}{3}$