## Calculus with Applications (10th Edition)

$$\frac{1}{{12}}$$
\eqalign{ & \int_0^1 {\int_y^{\sqrt y } {xdx} dy} \cr & = \int_0^1 {\left[ {\int_y^{\sqrt y } {xdx} } \right]dy} \cr & {\text{solve the inner integral by treating }}y{\text{ as a constant}} \cr & = \int_y^{\sqrt y } {xdx} = \left[ {\frac{{{x^2}}}{2}} \right]_y^{\sqrt y } \cr & {\text{evaluating the limits in the variable }}x \cr & = \frac{{{{\left( {\sqrt y } \right)}^2}}}{2} - \frac{{{{\left( y \right)}^2}}}{2} \cr & {\text{simplifying}} \cr & = \frac{1}{2}\left( {y - {y^2}} \right) \cr & \cr & \int_0^1 {\left[ {\int_y^{\sqrt y } {xdx} } \right]dy} = \frac{1}{2}\int_0^1 {\left( {y - {y^2}} \right)} dy \cr & {\text{integrating}} \cr & = \frac{1}{2}\left[ {\frac{{{y^2}}}{2} - \frac{{{y^3}}}{3}} \right]_0^1 \cr & = \frac{1}{2}\left( {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^3}}}{3}} \right) - \frac{1}{2}\left( {\frac{{{{\left( 0 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^3}}}{3}} \right) \cr & = \frac{1}{2}\left( {\frac{1}{6}} \right) \cr & = \frac{1}{{12}} \cr}