Answer
$$\frac{2}{{135}}\left( {{{42}^{5/2}} - {{24}^{5/2}} - {{39}^{5/2}} + {{21}^{5/2}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_3^4 {\int_2^5 {\sqrt {6x + 3y} dx} dy} \cr
& = \int_3^4 {\left[ {\int_2^5 {\sqrt {6x + 3y} dx} } \right]dy} \cr
& {\text{solve the inner integral}} \cr
& \int_2^5 {\sqrt {6x + 3y} dx} = \frac{1}{6}\int_2^5 {\sqrt {6x + 3y} \left( 6 \right)dx} \cr
& = \frac{1}{6}\left[ {\frac{{{{\left( {6x + 3y} \right)}^{3/2}}}}{{3/2}}} \right]_2^5 \cr
& = \frac{1}{9}\left[ {{{\left( {6x + 3y} \right)}^{3/2}}} \right]_2^5 \cr
& {\text{evaluate limits for }}x \cr
& = \frac{1}{9}\left[ {{{\left( {6\left( 5 \right) + 3y} \right)}^{3/2}} - {{\left( {6\left( 2 \right) + 3y} \right)}^{3/2}}} \right] \cr
& = \frac{1}{9}\left[ {{{\left( {30 + 3y} \right)}^{3/2}} - {{\left( {12 + 3y} \right)}^{3/2}}} \right] \cr
& {\text{then}} \cr
& \int_3^4 {\int_2^5 {\sqrt {6x + 3y} dx} dy} = \frac{1}{9}\int_3^4 {\left[ {{{\left( {30 + 3y} \right)}^{3/2}} - {{\left( {12 + 3y} \right)}^{3/2}}} \right]dy} \cr
& = \frac{1}{{27}}\int_3^4 {\left[ {{{\left( {30 + 3y} \right)}^{3/2}}\left( 3 \right) - {{\left( {12 + 3y} \right)}^{3/2}}\left( 3 \right)} \right]dy} \cr
& {\text{integrating by using the powe rule}} \cr
& = \frac{1}{{27}}\left[ {\frac{{{{\left( {30 + 3y} \right)}^{5/2}}}}{{5/2}} - \frac{{{{\left( {12 + 3y} \right)}^{5/2}}}}{{5/2}}} \right]_3^4 \cr
& = \frac{2}{{135}}\left[ {{{\left( {30 + 3y} \right)}^{5/2}} - {{\left( {12 + 3y} \right)}^{5/2}}} \right]_3^4 \cr
& evaluating \cr
& = \frac{2}{{135}}\left[ {{{\left( {30 + 3\left( 4 \right)} \right)}^{5/2}} - {{\left( {12 + 3\left( 4 \right)} \right)}^{5/2}}} \right] - \left[ {{{\left( {30 + 3\left( 3 \right)} \right)}^{5/2}} - {{\left( {12 + 3\left( 3 \right)} \right)}^{5/2}}} \right] \cr
& = \frac{2}{{135}}\left[ {{{\left( {42} \right)}^{5/2}} - {{\left( {24} \right)}^{5/2}}} \right] - \left[ {{{\left( {39} \right)}^{5/2}} - {{\left( {21} \right)}^{5/2}}} \right] \cr
& = \frac{2}{{135}}\left( {{{42}^{5/2}} - {{24}^{5/2}} - {{39}^{5/2}} + {{21}^{5/2}}} \right) \cr} $$
