Answer
$$\frac{{13}}{{80}}$$
Work Step by Step
$$\eqalign{
& \int_{1/4}^1 {\int_{{x^2}}^x {\sqrt {\frac{x}{y}} } dydx} \cr
& = \int_{1/4}^1 {\left[ {\int_{{x^2}}^x {\sqrt {\frac{x}{y}} } dy} \right]dx} \cr
& {\text{solve the inner integral and treat }}x{\text{ as a constant}} \cr
& \int_{{x^2}}^x {\sqrt {\frac{x}{y}} } dy = \int_{{x^2}}^x {\frac{{\sqrt x }}{{\sqrt y }}} dy \cr
& = \sqrt x \int_{{x^2}}^x {{y^{ - 1/2}}} dy \cr
& {\text{using the power rule }} \cr
& = \sqrt x \left( {\frac{{{y^{1/2}}}}{{1/2}}} \right)_{{x^2}}^x \cr
& = 2\sqrt x \left( {\sqrt y } \right)_{{x^2}}^x \cr
& {\text{evaluating the limits in the variable }}y \cr
& = 2\sqrt x \left( {\sqrt x - \sqrt {{x^2}} } \right) \cr
& {\text{simplifying}} \cr
& = 2\sqrt x \left( {\sqrt x - x} \right) \cr
& = 2x - 2{x^{3/2}} \cr
& {\text{then}} \cr
& \int_{1/4}^1 {\left[ {\int_{{x^2}}^x {\sqrt {\frac{x}{y}} } dy} \right]dx} = \int_{1/4}^1 {\left( {2x - 2{x^{3/2}}} \right)dx} \cr
& {\text{integrating}} \cr
& = \left( {{x^2} - \frac{{2{x^{5/2}}}}{{5/2}}} \right)_{1/4}^1 \cr
& = \left( {{x^2} - \frac{{4{x^{5/2}}}}{5}} \right)_{1/4}^1 \cr
& = \left( {{{\left( 1 \right)}^2} - \frac{{4{{\left( 1 \right)}^{5/2}}}}{5}} \right) - \left( {{{\left( {1/4} \right)}^2} - \frac{{4{{\left( {1/4} \right)}^{5/2}}}}{5}} \right) \cr
& = \frac{1}{5} - \frac{3}{{80}} \cr
& = \frac{{13}}{{80}} \cr} $$
