<s>[INST] Region area with double integral I need to find the area $\{(x, y) \in \mathbb{R}^2 : 0 < 2x < y < 3-x^2\}$ as the double integral $\int(\int dx) dy$.
I came up with:
$$\int_0^2\int_0^{\frac{1}{2}y}dxdy + \int_2^3\int_0^\sqrt{3-y}= \int_0^2\frac{1}{2}ydy + \int_2^3\sqrt{3-y}dy = \frac{1}{2}\frac{y^2}{2}\biggr\rvert^2_0 + \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 = \frac{1}{3}$$
But the solution gives $\frac{5}{3}$. I'm around this exercise a couple hours now and I can't seem to find my mistake.
 [/INST] Your answer is almost correct.
You made a single error when you took the antiderivative of the second integreal.
It should be $- \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $ not $+ \frac{2}{3}(3-y)^\frac{3}{2}\biggr\rvert_2^3 $
That will give you the correct result of $\frac{5}{3}$.
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