Respuesta :

caylus
Hello,
[tex]Area= \frac{16}{3}=2*( k*\sqrt{k}- \int\limits^{\sqrt{k}}_{0} {x^2} \, dx )\\\\ \dfrac{8}{3}=k ^\frac{3}{2}- \frac{1}{3}* [x^3]^{\sqrt{^k}}_0\\\\ 2* k ^\frac{3}{2}=8\\\\ k = \sqrt[3]{16}\\ [/tex]