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...
$\int_{-16}^{-4}\left(x^{2}-4 x\right) d x+\int_{-4}^{4}\left(4 x-x^{2}\right) d x$
B. $\int_{-16}^{0}\left(x^{2}-4 x\right) d x+\int_{0}^{4}\left(4 x-x^{2}\right) d x$
C. $\int_{-16}^{-4}\left(4 x-x^{2}\right) d x+\int_{-4}^{4}\left(x^{2}-4 x\right) d x$
D. $\int_{-16}^{-4}\left(x^{2}-4 x\right) d x+\int_{-4}^{4}\left(x^{2}-4 x\right) d x$
E. $\int_{-16}^{0}\left(4 x-x^{2}\right) d x+\int_{0}^{4}\left(x^{2}-4 x\right) d x$

Step 1 ：Suppose the function $f(x)$ is defined as $\left\{\begin{array}{l}x^{2}-4 x \text { for } x<-4 \\ 4 x-x^{2} \text { for } x \geq-4\end{array}\right.$

Step 2 ：We want to find the integral of $f(x)$ from -16 to 4, which is $\int_{-16}^{4} f(x) d x$

Step 3 ：We split the integral into two parts: from -16 to -4 and from -4 to 4

Step 4 ：For the first part, we use the function $x^{2}-4 x$, so the integral from -16 to -4 is $\int_{-16}^{-4}\left(x^{2}-4 x\right) d x$

Step 5 ：For the second part, we use the function $4 x-x^{2}$, so the integral from -4 to 4 is $\int_{-4}^{4}\left(4 x-x^{2}\right) d x$

Step 6 ：Therefore, the integral of $f(x)$ from -16 to 4 is $\int_{-16}^{-4}\left(x^{2}-4 x\right) d x+\int_{-4}^{4}\left(4 x-x^{2}\right) d x$

Step 7 ：The integral of the first part is 1824

Step 8 ：The integral of the second part is $-\frac{128}{3}$

Step 9 ：Adding these two integrals together, we get the final integral is $\frac{5344}{3}$

Step 10 ：Final Answer: The integral of the function $f(x)$ from -16 to 4 is $\boxed{\frac{5344}{3}}$