Suppose $f(x)=\{\begin{array}{l}{x}^2-4x\text{ for }x<-4\\ 4x-x^2\text{ for }x\ge -4\end{array}$ then
$${\int }_{-16}^{4}f(x)dx$$
is equal to...
${\int }_{-16}^{-4}(x^2-4x)dx+{\int }_{-4}^4(4x-x^2)dx$
B. ${\int }_{-16}^0(x^2-4x)dx+{\int }_0^4(4x-x^2)dx$
C. ${\int }_{-16}^{-4}(4x-x^2)dx+{\int }_{-4}^4(x^2-4x)dx$
D. ${\int }_{-16}^{-4}(x^2-4x)dx+{\int }_{-4}^4(x^2-4x)dx$
E. ${\int }_{-16}^0(4x-x^2)dx+{\int }_0^4(x^2-4x)dx$

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

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

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 $x^2-4x$, so the integral from -16 to -4 is ${\int }_{-16}^{-4}(x^2-4x)dx$

Step 5 ：For the second part, we use the function $4x-x^2$, so the integral from -4 to 4 is ${\int }_{-4}^4(4x-x^2)dx$

Step 6 ：Therefore, the integral of $f(x)$ from -16 to 4 is ${\int }_{-16}^{-4}(x^2-4x)dx+{\int }_{-4}^4(4x-x^2)dx$

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 $\boxed{{\displaystyle \frac{5344}{3}}}$