Answer
$$\int_{-2}^{2} f(x) d x = \frac{28}{3}$$
Work Step by Step
Given
$$\int_{-2}^{2} f(x) d x \quad \text { where } f(x)= \begin{cases}2 & \text { if }-2 \leqslant x \leqslant 0 \\ 4-x^{2} & \text { if } 0\lt x \leqslant 2\end{cases}$$
Since
\begin{aligned}
\int_{-2}^{2} f(x) d x &=\int_{-2}^{0} (2) d x +\int_{0}^{2} (4-x^2) d x \\
&=2x\bigg|_{-2}^0+ (4x-\frac{1}{3}x^3)\bigg|_0^2\\
&= (0)-(-4)+ [(8-\frac{8}{3}) - (0) ]\\
&= \frac{28}{3}
\end{aligned}