# How to solve <msubsup> &#x222B;<!-- ∫ --> 0 1 </msubsup> <msubsup> &#x222B;<!

How to solve ${\int }_{0}^{1}{\int }_{0}^{1}\sqrt{1-x}\sqrt{1-y}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}dxdy$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

stormsteghj
Since $\sqrt{1-x}=\left(1-x{\right)}^{1/2}$, an antiderivative of which is $-\frac{2}{3}\left(1-x{\right)}^{3/2}$, you have
${\int }_{0}^{1}\sqrt{1-x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x={\left[-\frac{2}{3}\left(1-x{\right)}^{3/2}\right]}_{x=0}^{x=1}=\frac{2}{3}$
and therefore
${\int }_{0}^{1}\sqrt{1-x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x{\int }_{0}^{1}\sqrt{1-y}\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}y={\left(\frac{2}{3}\right)}^{2}=\frac{4}{9}.$