## Calculus with Applications (10th Edition)

$$\frac{{128}}{9}$$
\eqalign{ & \int_0^4 {\int_0^x {\sqrt {xy} } dydx} \cr & = \int_0^4 {\int_0^x {\sqrt x \sqrt y } dydx} \cr & = \int_0^4 {\left[ {\int_0^x {\sqrt x \sqrt y } dy} \right]dx} \cr & {\text{solve the inner integral, treating }}x{\text{ as a constant}} \cr & = \sqrt x \int_0^x {\sqrt y } dy \cr & = \sqrt x \int_0^x {{y^{1/2}}} dy \cr & {\text{integrate by using the power rule }} \cr & = \sqrt x \left[ {\frac{{2{y^{3/2}}}}{3}} \right]_0^x \cr & {\text{evaluating the limits in the variable }}y \cr & = \frac{{2\sqrt x }}{3}\left( {{{\left( x \right)}^{3/2}} - {{\left( 0 \right)}^{3/2}}} \right) \cr & {\text{simplifying}} \cr & = \frac{{2\sqrt x }}{3}\left( {{x^{3/2}}} \right) \cr & = \frac{{2\sqrt x }}{3}\left( {x\sqrt x } \right) \cr & = \frac{2}{3}{x^2} \cr & \cr & \int_0^4 {\left[ {\int_0^x {\sqrt x \sqrt y } dy} \right]dx} = \int_0^4 {\left( {\frac{2}{3}{x^2}} \right)dx} \cr & = \frac{2}{3}\int_0^4 {{x^2}dx} \cr & {\text{integrating by the power rule}} \cr & = \frac{2}{3}\left( {\frac{{{x^3}}}{3}} \right)_0^4 \cr & = \frac{2}{9}\left( {{x^3}} \right)_0^4 \cr & {\text{evaluate}} \cr & = \frac{2}{9}\left( {{4^3} - {0^3}} \right) \cr & {\text{simplifying}} \cr & = \frac{{128}}{9} \cr}