Answer
$$ - \frac{6}{5}$$
Work Step by Step
$$\eqalign{
& \iint\limits_R {\left( { - 2y} \right)}dA \cr
& {\text{Let the region }}R:y = 4 - {x^2},{\text{ }}y = 4 - x \cr
& R = \left\{ {\left( {x,y} \right):4 - x \leqslant y \leqslant 4 - {x^2},{\text{ }}0 \leqslant x \leqslant 1} \right\} \cr
& {\text{Therefore,}} \cr
& \iint\limits_R {\left( { - 2y} \right)}dA = \int_0^1 {\int_{4 - x}^{4 - {x^2}} {\left( { - 2y} \right)dy} dx} \cr
& = \int_0^1 {\left[ {\int_{4 - x}^{4 - {x^2}} {\left( { - 2y} \right)dy} } \right]dx} \cr
& {\text{Integrate with respect to }}y \cr
& = \int_0^1 {\left[ { - {y^2}} \right]_{4 - x}^{4 - {x^2}}dx} \cr
& = - \int_0^1 {\left[ {{{\left( {4 - {x^2}} \right)}^2} - {{\left( {4 - x} \right)}^2}} \right]dx} \cr
& = - \int_0^1 {\left( {16 - 8{x^2} + {x^4} - 16 + 8x - {x^2}} \right)dx} \cr
& = - \int_0^1 {\left( {{x^4} - 9{x^2} + 8x} \right)dx} \cr
& {\text{Integrating}} \cr
& = - \left[ {\frac{{{x^5}}}{5} - 3{x^3} + 4{x^2}} \right]_0^1 \cr
& = - \left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - 3{{\left( 1 \right)}^3} + 4{{\left( 1 \right)}^2}} \right] + \left[ 0 \right] \cr
& = - \frac{6}{5} \cr} $$
