Answer
$\dfrac{4}{3}$
Work Step by Step
Consider $Area; A=\int_{-2}^{0} \int_{ 2x+4}^{4-x^2} dy \ dx$
or, $=\int_{-2}^{0} (-x^2 -2x) \ dx $
or, $=- \int_{-2}^{0} x^2 \ dx - 2 \int_{-2}^{0} x \ dx $
or, $=-[ \dfrac{x^3}{3}]_{-2}^0 - [ x^2]_{-2}^0 $
or, $=-[ 0-\dfrac{(-2)^3}{3}] - [0- (-2)^2] $
or, $=\dfrac{4}{3}$
