Answer
$$\frac{125}{6}$$
Work Step by Step
Using the rules of integrals, we obtain:
Area $=\int_{-4}^{1} \int_{3 y-4}^{-y^{2}} d x d y=\iint_{R} d A$
$=\int_{-4}^{1}[x]_{3 y-4}^{-y^{2}} d y$
$=\int_{-4}^{1}4-y^{2}-3 y+d y$
$=\left[-\frac{3 y^{2}}{2}-\frac{y^{3}}{3}+4 y\right]_{-4}^{1}$
$=\left[-\frac{1^{3}}{3}-\frac{3 \cdot 1^{2}}{2}+4 \cdot 1\right]-\left[-\frac{(-4)^{3}}{3}-\frac{3(-4)^{2}}{2}+4(-4)\right]$
$\begin{aligned}{=\left[-\frac{1^{3}}{3}-\frac{3 \cdot 1^{2}}{2}+4 \cdot 1\right]-\left[-\frac{(-4)^{3}}{3}-\frac{3(-4)^{2}}{2}+4(-4)\right]}{=\left[-\frac{1}{3}-\frac{3}{2}+4\right]-\left[\frac{64}{3}-24-16\right]} \\ &=-\frac{64}{3}+24+16-\frac{1}{3}-\frac{3}{2}+4=\frac{125}{6} \end{aligned}$
