Question about continuous joint pdf of two random variables $\text{X and Y are a pair of jointly continuous random variables:}$

$f_{X,Y}(x, y) = \begin{cases}
       \text{c,} &\text{the shaded region}\\
       \text{0,} &\text{elsewhere}\\
     \end{cases}$
$\text{a- find the value of c.}$
$\text{b- find the marginal pdf of X and Y.}$
$\text{c- find E[X|Y=1/4] and Var[X|Y=1/4]}.$
$\text{d- find the conditional PDF of X knowing that Y=3/4.}$
$\text{what I did:}$
for the first question, we know that the function is constant thus we can simply multiply the shaded surface and c and this value will equal 1.
$$c\times S = \frac{c}{2} = 1$$
$$c = 2$$
for the marginal pdf of each variable:
the definition domain is:
$$\Delta(x,y)=\text{{0⩽x⩽1 and 0⩽y⩽1, and [0.5-x⩽y⩽1-x, or 3/2-x⩽y]}}$$
$$f_X(x) = \int^{1/2-x}_{1-x}cdy + \int^{3/2-x}_{1}cdy$$
$$f_X(x) = 2x$$
$$f_Y(y) = \int^{1-y}_{1/2-y}cdx + \int^{1}_{3/2-y}cdx$$
$$f_Y(y) = 2y$$
however when I tried to compute the expectation I got a negative value:
$$E[X|Y] = \int^{1-y}_{1/2-y}x \times \frac{f(x,y)}{f(y)}dx + \int^{1}_{3/2-y}x \times \frac{f(x,y)}{f(y)}dx$$
$$E[X|Y] = \int^{1-y}_{1/2-y}\frac{x}{y}dx + \int^{1}_{3/2-y}\frac{x}{y}dx$$
$$E[X|Y] = \frac{8y - 2 - 4y}{8y}$$
$$E[X|Y=1/4] = -0.5$$
I also computed in the same way the variance and found that it was negative [obviously wrong], thus I would be sincerely grateful to know where I made my mistake.

There is a support mistake when deriving the marginal densities (and again in constructing the conditional expectations) \begin{align}
f_Y(y) &\propto\mathbb I_{0\le y\le 1/2}\int^{1-y}_{1/2-y}\text dx 
+ \mathbb I_{1\ge y\ge 1/2}\int^{1-y}_{0}\text dx
 + \mathbb I_{1\ge y\ge 1/2}\int^{1}_{3/2-y}\\
&= \begin{cases}{\left[(1-y)-0+1-(3/2-y)
\right] \text{ if }1\ge y\ge 1/2\\
\left[(1-y)-(1/2-y)\right] \text{ otherwise}
}\end{cases}\\
&= \begin{cases}{\dfrac{1}{2}\text{ if }1\ge y\ge 1/2\\
\dfrac{1}{2}\text{ otherwise}
}\end{cases}\\
&\propto \mathbb I_{(0,1)}(y)
\end{align}
meaning that both $f_X$ and $f_Y$ are densities for the Uniform $\mathcal U(0,1)$ distribution.
