Marginal density of random vector distributed on trapezoid (bounds of integral)

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Let $R \subset \mathbb{R}^2$ be a right-angled trapezoid with $(0,0), (2,0), (2,1), (1,1)$ as corners. Let $(X,Y)$ be a random vector uniformly distributed on $R$.

Obviously, the the joint density is $$f_{X,Y}(s_1,s_2) = \frac{1}{area(R)} = \frac{2}{3} , (s_1,s_2) \in R$$

Since $area(R) = \frac{1}{2}h(a+b) = \frac{3}{2}$.

I now want to calculate the marginal density $$f_Y(s_2) = \int_{-\infty}^\infty f_{X,Y}(s_1,s_2) ds_1 = \int_{-\infty}^\infty f_{X,Y}(s_1,s_2) ds_1 = \int_{-\infty}^\infty \frac{2}{3} ds_1$$

I wonder about the bounds of the integral. If I draw the trapezoid I can see that $0 \leq s_2 \leq 1$ and $0 \leq s_1 \leq 2$. So, I guess the upper bound is $2$. But what about the lower bound... it cannot be $0$, right? Shouldn't the bounds to calculate the marginal density always be in terms of the opposite variable, in this case $s_2$?

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For given $s_2\in(0,1)$ you should integrate by $s_1$ from $s_1=s_2$ to $s_1=2$ as indicated on the following picture:

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