Evaluate $\int_{-1}^{1}\int_{x}^{2x-1}dydx$ 
$$\int_{-1}^{1}\int_{x}^{2x-1}dydx$$

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My attempt:
$$I_1:=\int_{-1}^{1}\int_{x}^{2x-1}dydx$$
$$=\int_{-1}^{1}\bigg[\int_{x}^{2x-1}dy\bigg]dx$$
$$=\int_{-1}^{1}\bigg(x-1\bigg)dx=\boxed{\color{red}{-2}}$$
Now I need to evaluate with changing the limis:
$$I_2:=\int_{-3}^{-1}\int_{-1}^{\frac{y+1}{2}}dxdy+\int_{-1}^{1}\int_{y}^{\frac{y+1}{2}}dxdy$$

1) Is it ok that before changing the limits the area is negative?
2) The new limits that I have done are correct?


After changing the limits:

$$I_2=\int_{-3}^{-1}\int_{-1}^{\frac{y+1}{2}}dxdy+\int_{-1}^{1}\int_{y}^{\frac{y+1}{2}}dxdy$$

$$=\underbrace{\int_{-3}^{-1}\bigg(\frac{y+1}{2}+1\bigg)}_{=\color{blue}1}dy+\underbrace{\int_{-1}^{1}\bigg(\frac{y+1}{2}-y\bigg)}_{=\color{blue}1}dy$$
$$$$
$$\Longrightarrow=1+1=2$$
