Answer
$$56$$
Work Step by Step
$$\eqalign{
& \int_0^4 {\int_0^1 {\int_0^2 {\left( {2x + y + 4z} \right)dy} } dz} dx \cr
& \int_0^4 {\int_0^1 {\left[ {\int_0^2 {\left( {2x + y + 4z} \right)dy} } \right]} dz} dx \cr
& {\text{Integrate with respect to }}y \cr
& \int_0^2 {\left( {2x + y + 4z} \right)dy} = \left[ {2xy + \frac{1}{2}{y^2} + 4yz} \right]_0^2 \cr
& = \left[ {2x\left( 2 \right) + \frac{1}{2}{{\left( 2 \right)}^2} + 4\left( 2 \right)z} \right] - \left[ {2x\left( 0 \right) + \frac{1}{2}{{\left( 0 \right)}^2} + 4\left( 0 \right)z} \right] \cr
& = 4x + 2 + 8z \cr
& = \int_0^4 {\int_0^1 {\left( {4x + 2 + 8z} \right)} dz} dx \cr
& = \int_0^4 {\left[ {\int_0^1 {\left( {4x + 2 + 8z} \right)} dz} \right]} dx \cr
& {\text{Integrate with respect to }}z \cr
& \int_0^1 {\left( {4x + 2 + 8z} \right)} dz = \left[ {4xz + 2z + 4{z^2}} \right]_0^1 \cr
& = \left[ {4x\left( 1 \right) + 2\left( 1 \right) + 4{{\left( 1 \right)}^2}} \right] - \left[ {4x\left( 0 \right) + 2\left( 0 \right) + 4{{\left( 0 \right)}^2}} \right] \cr
& = 4x + 6 \cr
& = \int_0^4 {\left( {4x + 6} \right)} dx \cr
& {\text{Integrate}} \cr
& {\text{ = }}\left[ {2{x^2} + 6x} \right]_0^4 \cr
& = 2{\left( 4 \right)^2} + 6\left( 4 \right) \cr
& = 56 \cr} $$
