Question

calcular la integral doble

Answer 1

               $\displaystyle I=\iint\limits_R\left(3x-4x\sqrt{xy}\right)dA\\ \\ \\ I=\int_{0}^{4}\int_{0}^{9}\left(3x-4x\sqrt{xy}\right)~dy~dx\\ \\ \\ I=\int_{0}^{4}x\int_{0}^{9}\left(3-4\sqrt{xy}\right)~dy~dx\\ \\ \\ I=\int_{0}^{4}x\left.\left(3y-\dfrac{8y\sqrt{xy}}{3}\right)\right|_{y=0}^{y=9}dx\\ \\ \\ I=\int_{0}^{4}x(27-72\sqrt{x})dx\\ \\ \\ I=\left.\left(\dfrac{27}{2}x^2-\dfrac{144x^2\sqrt{x}}{5}\right)\right|_{x=0}^{x=4}\\ \\ \\ \\ I=27\times 8-\dfrac{144}{5}\times 32=-\dfrac{3528}{5}$