Answer
$$A = \frac{4}{{15}}$$
Work Step by Step
$$\eqalign{
& y = \sqrt x ,\,\,\,\,\,\,y = x\sqrt x \cr
& {\text{Find the points of intersection of the graphs equating the functions}} \cr
& \sqrt x = x\sqrt x \cr
& {x^{1/2}} = {x^{3/2}} \cr
& {x^{1/2}} - {x^{3/2}} = 0 \cr
& \left( {1 - x} \right){x^{1/2}} = 0 \cr
& x = 0,\,\,x = 1 \cr
& {\text{Then with the critical value we have the interval }}\left( {0,1} \right) \cr
& {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^{1/2}} > {x^{3/2}} \cr
& \cr
& {\text{The area between the curves is given by}}: \cr
& A = \int_0^1 {\left( {{x^{1/2}} - {x^{3/2}}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\frac{{2{x^{3/2}}}}{3} - \frac{{2{x^{5/2}}}}{5}} \right)_0^1 \cr
& A = \left( {\frac{{2{{\left( 1 \right)}^{3/2}}}}{3} - \frac{{2{{\left( 1 \right)}^{5/2}}}}{5}} \right) - \left( {\frac{{2{{\left( 0 \right)}^{3/2}}}}{3} - \frac{{2{{\left( 0 \right)}^{5/2}}}}{5}} \right) \cr
& A = \frac{2}{3} - \frac{2}{5} \cr
& A = \frac{4}{{15}} \cr} $$
