If I want to find the $c$ of a $PDF$ when it's given:
$f_{X,Y}\left(x,y\right)=c\:\:\:\:\left(The\:area\:in\:blue\right),\:otherwise:\:0$
I try to do that:
$$\int _{\frac{1}{2}}^1\:\int _{-x+\frac{3}{2}}^1\:c\,dy\,dx+\int _0^1\:\int _{-x+\frac{1}{2}}^{-x+1}\:c\,dy\,dx=1$$
But I got $c=\frac{8}{5}$ and the result is $c=2$.
(I know we can calculate it with by calculating the area of the triangles but I am wondering to know why my way is not working)
On
$$ \begin{aligned} \iint\limits_{\text{square}}f(x, y)dxdy &= \iint\limits_{\text{white triangle}}0dxdy+ \iint\limits_{\text{blue trapezoid}}cdxdy+ \iint\limits_{\text{white trapezoid}}0dxdy+ \iint\limits_{\text{blue triangle}}cdxdy = \\ &= c\left(\iint\limits_{\text{blue trapezoid}}dxdy+ \iint\limits_{\text{blue triangle}}dxdy\right) = \\ &= \left| \begin{aligned} &\text{Integral over the blue trapezoid can be represented as a sum of two integrals:}\\ &\qquad\text{integral over a blue parallelogram (left to the }x = \frac{1}{2}\text{) and }\\ &\qquad\text{integral over a blue triangle (right to the }x = \frac{1}{2}\text{)} \end{aligned} \right| = \\ &= c\left(\int\limits_0^{\frac{1}{2}}dx\int\limits_{-x + \frac{1}{2}}^{-x + 1}dy + \int\limits_{\frac{1}{2}}^1dx\int\limits_{0}^{-x + 1}dy + \int\limits_{\frac{1}{2}}^1dx\int\limits_{-x + \frac{3}{2}}^1dy \right) = \\ &= c\left(\int\limits_0^\frac{1}{2}\frac{1}{2}dx + \int\limits_\frac{1}{2}^1(-x+1)dx + \int_\frac{1}{2}^1\left(x-\frac{1}{2}\right)dx\right) = \\ &= c\left(\frac{1}{4} - \left.\frac{(-x+1)^2}{2}\right|_\frac{1}{2}^1+ \left.\frac{\left(x-\frac{1}{2}\right)^2}{2}\right|_\frac{1}{2}^1\right) = \\ &= c\left(\frac{1}{4} + \frac{1}{8} + \frac{1}{8}\right) = c\cdot\frac{1}{2} = 1 \Rightarrow c = 2 \end{aligned} $$
You need to calculate like this:
$$\int _0^{\frac{1}{2}}\:\int _{\frac{1}{2}-x}^{-x+1}\:cdydx+\int _{\frac{1}{2}}^1\int _0^{-x+1}\:\:cdydx+\int _{\frac{1}{2}}^1\int _{-x+\frac{3}{2}}^1\:\:cdydx=1$$
And then you'll get: $c=2$