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I'm given a $$f_{X,Y}(x,y) = \begin{cases} cx, & \text{x > 0, y > 0, 1}\ \leq \ x+y \ \leq 2, \\ 0, & \text{elsewhere.} \end{cases}$$ and trying to find a the constant $c$.

I've set the x range to $$0 < x < 2$$ and y range to $$1-x < y < 2-x$$ but I'm confused as if this should be $0 < y < 2-x$ instead

The answer said $$c\int_{0}^1\int_{1-x}^{2-x}xdydx + c\int_{1}^2\int_{0}^{2-x}xdydx$$ and continue the calculation from here.

Why should the equation be formed in this way?

Xenotion
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  • You don't have an equation yet. The point is that the integral of the density function is equal to $1$, which gives you an equation in the unknown $c$. –  Jun 03 '22 at 13:44
  • @user1046533 I think my question was more on the line of why do the integrals have to be separated into when 0 < x <= 1 and 1 < x < 2 and match the y integral values accoriding to that – Xenotion Jun 03 '22 at 13:53
  • The inequalities need to be considered altogether instead of just a single one. –  Jun 03 '22 at 14:16

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So you need to calculate the double integral $$ \iint_Ef(x,y)\;dxdy\quad E=\{(x,y):x>0,y>0,1\le x+y\le 2\} $$ The region $E$ is not a "normal domain" in $\mathbf{R}^2$. So in order to write the integral as iterated integrals, you need to split the region into two parts so that they are both normal domains. Your solution divide the region as follows:

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