Answer
$$8$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\int_0^1 {\left( {3{x^2} + 4{y^3}} \right)} } dydx \cr
& = \int_1^2 {\left[ {\int_0^1 {\left( {3{x^2} + 4{y^3}} \right)} dy} \right]} dx \cr
& {\text{solve the inner integral}}{\text{, treat }}x{\text{ as a constant}} \cr
& \int_0^1 {\left( {3{x^2} + 4{y^3}} \right)} dy = \left[ {3{x^2}\left( y \right) + 4\left( {\frac{{{y^4}}}{4}} \right)} \right]_0^1 \cr
& = \left[ {3{x^2}y + {y^4}} \right]_0^1 \cr
& {\text{evaluating the limits in the variable }}y \cr
& = \left[ {3{x^2}\left( 1 \right) + {{\left( 1 \right)}^4}} \right] - \left[ {3{x^2}\left( 0 \right) + {{\left( 0 \right)}^4}} \right] \cr
& {\text{simplifying}} \cr
& = 3{x^2} + 1 \cr
& \cr
& \int_1^2 {\left[ {\int_0^1 {\left( {3{x^2} + 4{y^3}} \right)} dy} \right]} dx = \int_1^2 {\left( {3{x^2} + 1} \right)} dx \cr
& {\text{integrating}} \cr
& = \left( {{x^3} + x} \right)_1^2 \cr
& {\text{evaluate}} \cr
& = \left( {{2^3} + 2} \right) - \left( {{1^3} + 1} \right) \cr
& = 10 - 2 \cr
& = 8 \cr} $$
