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