I don't understand this topic honestly.
I thought there is supposed to be two types of ranges for x and y. $$\therefore \space \mathsf{for} \space \space x^2-1 \leq 3 \space ; \space x \in [-2,2] \space \space \mathsf{and} \space \space y \in [x^2-1,3]$$ $$\therefore \space \mathsf{for} \space \space x^2-1 \leq 2x+2 \space; \space x \in [-1,3] \space \space \mathsf{and} \space \space y \in [x^2-1,2x+2]$$
Then I drew the graphs of both cases separately:
And constructed what I understood from this:
$$\displaystyle \int \limits^2_{-2} \int \limits^3_{(x^2-1)} (4-x^2) \mathsf{d}y \mathsf{d}x + \int \limits^3_{-1} \int \limits^{2x+2}_{x^2-1} (2x+3-x^2) \mathsf{d}y \mathsf{d}x$$
Apparently, from what I've seen, my way is not correct. What is the correct way and reason to find this double integral?
Combining those two graphs, you can notice that the area is divided into two sections:
from -1 to $\frac{1}{2}$ that is lower constrained by $x^2-1$ and upper constrained by $2x+2$,
from $\frac{1}{2}$ to 2 that is lower constrained by $x^2-1$ and upper constrained by $3$.
Can you continue on your own from there?