## Thomas' Calculus 13th Edition

$$\dfrac{4}{3}$$
We calculate the integral as follows: Area; $$A=\int_{-2}^{0} \int_{ 2x+4}^{4-x^2} dy dx \\=\int_{-2}^{0} (-x^2 -2x) dx \\=- \int_{-2}^{0} x^2 \ dx - 2 \int_{-2}^{0} x \ dx \\=-\dfrac{1}{3} \times [ x^3]_{-2}^0 - [ x^2]_{-2}^0 \\=-[ 0-\dfrac{(-2)^3}{3}] - [0- (-2)^2] \\=\dfrac{4}{3}$$