Answer
$$\frac{2}{{45}}\left[ {{{\left( {24} \right)}^{5/2}} - {{\left( {21} \right)}^{5/2}} - {{\left( {15} \right)}^{5/2}} + {{\left( {12} \right)}^{5/2}}} \right]$$
Work Step by Step
$$\eqalign{
& \int_0^3 {\int_4^5 {x\sqrt {{x^2} + 3y} } } dydx \cr
& = \int_0^3 {\left[ {\int_4^5 {x\sqrt {{x^2} + 3y} } dy} \right]} dx \cr
& {\text{solve the inner integral}} \cr
& = \frac{1}{3}x\int_4^5 {{{\left( {{x^2} + 3y} \right)}^{1/2}}\left( 3 \right)dy} \cr
& {\text{using the power rule }} \cr
& = \frac{1}{3}x\left( {\frac{{{{\left( {{x^2} + 3y} \right)}^{3/2}}}}{{3/2}}} \right)_4^5 \cr
& {\text{then}} \cr
& = \frac{2}{9}x\left( {{{\left( {{x^2} + 3y} \right)}^{3/2}}} \right)_4^5 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \frac{2}{9}x\left( {{{\left( {{x^2} + 3\left( 5 \right)} \right)}^{3/2}} - {{\left( {{x^2} + 3\left( 4 \right)} \right)}^{3/2}}} \right) \cr
& {\text{simplifying}} \cr
& = \frac{2}{9}x\left( {{{\left( {{x^2} + 15} \right)}^{3/2}} - {{\left( {{x^2} + 12} \right)}^{3/2}}} \right) \cr
& = \int_0^3 {\frac{2}{9}x\left( {{{\left( {{x^2} + 15} \right)}^{3/2}} - {{\left( {{x^2} + 12} \right)}^{3/2}}} \right)dx} \cr
& = \frac{1}{9}\int_0^3 {{{\left( {{x^2} + 15} \right)}^{3/2}}} \left( {2x} \right)dx - \frac{1}{9}\int_0^3 {{{\left( {{x^2} + 12} \right)}^{3/2}}} \left( {2x} \right)dx \cr
& {\text{integrating}} \cr
& = \frac{1}{9}\left[ {\frac{{2{{\left( {{x^2} + 15} \right)}^{5/2}}}}{5}} \right]_0^3 - \frac{1}{9}\left[ {\frac{{2{{\left( {{x^2} + 12} \right)}^{5/2}}}}{5}} \right]_0^3 \cr
& = \frac{2}{{45}}\left[ {{{\left( {{x^2} + 15} \right)}^{5/2}} - {{\left( {{x^2} + 12} \right)}^{5/2}}} \right]_0^3 \cr
& evaluating \cr
& = \frac{2}{{45}}\left[ {{{\left( {{3^2} + 15} \right)}^{5/2}} - {{\left( {{3^2} + 12} \right)}^{5/2}}} \right] - \frac{2}{{45}}\left[ {{{\left( {{0^2} + 15} \right)}^{5/2}} - {{\left( {{0^2} + 12} \right)}^{5/2}}} \right] \cr
& = \frac{2}{{45}}\left[ {{{\left( {24} \right)}^{5/2}} - {{\left( {21} \right)}^{5/2}} - {{\left( {15} \right)}^{5/2}} + {{\left( {12} \right)}^{5/2}}} \right] \cr} $$
