I sketched out the graph to get this figure but I can't seem to find the area of the shaded region... would one Y = 4 and the other Y = 8? 
I, in all honesty, am so flabbergasted with this question, any guidance will be much appreciated. I got 28/3 for the first integral (0 to 4) and 292/3 for the second integral (4 to 16)... I feel like I'm doing this wrong.
Okay, I'm going to be honest, can someone please solve this for me? I got 320/3 as my final answer and yet it's still wrong, I'm just about given up at this point, it's literally my last question of this semester.
We want,
$$\left(\int_{0}^{4} \sqrt{x}\,\,dx-\int_{0}^{4}\dfrac12x\,\,dx\right)+\left(\int_{4}^{16}\dfrac12x\,\,dx-\int_{4}^{16}\sqrt{x}\,\,dx\right)$$
where the first term two terms give the area from $0$ to $4$ and the second term gives it from $4$ to $16$.
The expression becomes: $\left(\dfrac23x^{3/2}-\dfrac14x^2\right)\left.\right|_0^4+\left(\dfrac14x^2-\dfrac23x^{3/2}\right)\left.\right|_4^{16}$