Answer
$\dfrac{8}{3}$
Work Step by Step
Re-arrange the given integral as follows:
$\int_{0}^{4}\int_{1}^{2}(\dfrac{\sqrt x}{y^2}) dy dx=\int_{1}^{2} (\dfrac{2}{3}x^{3/2})_0^4 y^{-2} dy$
This implies that
$\int_{1}^{2} (\dfrac{2}{3}x^{3/2})_0^4 y^{-2} dy=(\dfrac{16}{3})(-y^{-1})]_{1}^{2}$
Hence, $(\dfrac{16}{3})[1-\dfrac{1}{2}]=\dfrac{8}{3}$
