Answer
$$\frac{{45}}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {\int_{1 - x}^{8 - x} {{{\left( {x + y} \right)}^{1/3}}} dydx} \cr
& \int_0^2 {\left[ {\int_{1 - x}^{8 - x} {{{\left( {x + y} \right)}^{1/3}}} dy} \right]dx} \cr
& {\text{Evaluate inner integral}} \cr
& \int_{1 - x}^{8 - x} {{{\left( {x + y} \right)}^{1/3}}} dy = \left[ {\frac{{{{\left( {x + y} \right)}^{4/3}}}}{{4/3}}} \right]_{y = 1 - x}^{y = 8 - x} \cr
& = \frac{3}{4}\left[ {{{\left( {x + y} \right)}^{4/3}}} \right]_{y = 1 - x}^{y = 8 - x} \cr
& = \frac{3}{4}\left[ {{{\left( {x + 8 - x} \right)}^{4/3}} - {{\left( {x + 1 - x} \right)}^{4/3}}} \right] \cr
& = \frac{3}{4}\left[ {{{\left( 8 \right)}^{4/3}} - {{\left( 1 \right)}^{4/3}}} \right] \cr
& = \frac{3}{4}\left( {15} \right) \cr
& = \frac{{45}}{4} \cr
& {\text{Therefore,}} \cr
& \int_0^2 {\left[ {\int_{1 - x}^{8 - x} {{{\left( {x + y} \right)}^{1/3}}} dy} \right]dx} = \int_0^2 {\frac{{45}}{4}dx} \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{{45}}{4}\left[ x \right]_0^2 \cr
& = \frac{{45}}{4}\left[ {2 - 0} \right] \cr
& = \frac{{45}}{2} \cr} $$
