Suppose $f(x)=\left\{\begin{array}{l}x^{2}-4 x \text { for } x< -4 \\ 4 x-x^{2} \text { for } x \geq-4\end{array}\right.$ then
\[
\int_{-16}^{4} f(x) d x
\]
is equal to...
A. $\int_{-16}^{-4}\left(4 x-x^{2}\right) d x+\int_{-4}^{4}\left(x^{2}-4 x\right) d x$
B. $\int_{-16}^{-4}\left(x^{2}-4 x\right) d x+\int_{-4}^{4}\left(4 x-x^{2}\right) d x$
C. $\int_{-16}^{-4}\left(x^{2}-4 x\right) d x+\int_{-4}^{4}\left(x^{2}-4 x\right) d x$
D. $\int_{-16}^{0}\left(4 x-x^{2}\right) d x+\int_{0}^{4}\left(x^{2}-4 x\right) d x$
$\int_{-16}^{0}\left(x^{2}-4 x\right) d x+\int_{0}^{4}\left(4 x-x^{2}\right) d x$

Step 1 ：Given the piecewise function $f(x)=\left\{\begin{array}{l}x^{2}-4 x \text { for } x<-4 \\ 4 x-x^{2} \text { for } x \geq-4\end{array}\right.$, we are asked to find the integral of $f(x)$ from $-16$ to $4$.

Step 2 ：To do this, we need to split the integral at $x=-4$, where the function changes.

Step 3 ：We integrate $x^{2}-4 x$ from $-16$ to $-4$ and $4 x-x^{2}$ from $-4$ to $4$.

Step 4 ：The integral of $x^{2}-4 x$ from $-16$ to $-4$ is 1824.

Step 5 ：The integral of $4 x-x^{2}$ from $-4$ to $4$ is $-\frac{128}{3}$.

Step 6 ：Adding these two results together, we find that the integral of $f(x)$ from $-16$ to $4$ is $\frac{5344}{3}$.

Step 7 ：Thus, the final answer is \(\boxed{\frac{5344}{3}}\).