Answer
$\dfrac{4}{35}$
Work Step by Step
$\int^1_0 \int^{\sqrt{y}}_y \sqrt{x} \ dx \ dy =\dfrac{2}{3} \times \int^1_0 [x^{3/2}]^{\sqrt{y}}_y \ dy $
or, $=\dfrac{2}{3} \int^1_0 (y^{3/4}-y^{3/2})dy$
or, $=\dfrac{2}{3} [\dfrac{4}{7}y^{7/4}-\dfrac{2}{5}y^{5/2}]^1_0$
or, $=\dfrac{2}{3}(\dfrac{4}{7}-\dfrac{2}{5})$
or, $=\dfrac{4}{35}$
Next, $\int^1_0 \int^x_{x^2}\sqrt{x} \ dy \ dx =\int^1_0 x^{1/2}(x-x^2)dx $
or, $=\int^1_0 (x^{3/2}-x^{5/2})dx $
or, $=[\dfrac{2}{5}x^{5/2}-\dfrac{2}{7}x^{7/2}]^1_0$
or, $=\dfrac{2}{5}-\dfrac{2}{7}$
or, $=\dfrac{4}{35}$
