Find $E(X \mid Y=X+1/2)$ for $(X,Y)$ with given PDF suppose    f( x , y ) = x + y  
$ 0 < x < 1$, $ 0  <y < 1 $. 
Find $E(X \mid Y=X+\frac{1}{2})$
So far, what I got,
\begin{align}
f_{X \mid Y=X+\frac{1}{2}}
   & = \frac{\int_{x}^1 (x+y) \mathbb{1}\,dy}
            {\int_{0}^1 \int_{x}^1 (x+y) \mathbb{1}dy\,dx} \\
     & = \frac{x+   \frac{1}{2} -\frac{3x^2}{2}}{\int_{0}^1 (x+\frac{1}{2} -\frac{3x^2}{2}) \,dx} \\
   & = 2x+1-3x^2
\end{align}
$$
E(X \mid Y=X+\frac{1}{2}) = \int_{0}^1 xf_{X\mid Y=X+(1/2)}) \, dx
                = \int_{0}^1 (x)(2x+1-3x^2) \, dx
                = \left. \frac{2x^3}{3} +\frac{x^2}{2}-\frac{3x^4}{4} \right|_{0}^1 = \frac{5}{12}
$$
I'm still confused, this expectation and conditional distribution..
is it correct?

Note: $\mathsf E(X\mid y=x+1/2) = \mathsf E(X)\; [y-x=1/2]$ because lowercase $x,y$ are constant values, not random variables.  I'm assuming this is not actually what you want.

You have $f_{X,Y}(x,y)= (x+ y ) \;\big[x\in(0;1), y\in(0;1)\big]$.
So the support is the unit square.
I'm assuming you actually wish to evaluate: $\mathsf E\big(X \;\big\vert\; Y=X+\tfrac 1 2\big)$
That is the expectation of X when $(X,Y)=(X,X+\tfrac 12)$, which lies on the line-segement which runs from point $(0,\tfrac 12)$ to point $(\tfrac 1 2, 1)$. (AKA line $Y=X+\tfrac 1 2$ in the unit square.)
So: $$\begin{align}\mathsf E\big(X \;\big\vert\; Y=X+\tfrac 1 2\big) & = \int_0^{1/2} x f_{X\mid Y=X+1/2}(x)\operatorname d x 
\\[1ex] 
& =  \frac{\int_0^{1/2}x\,f_{X,Y}(x,x+\tfrac 1 2)\operatorname d x}{\int_0^{1/2}f_{X,Y}(z,z+\tfrac 1 2)\operatorname d z}
\\[1ex] 
& =  \frac{\int_0^{1/2} (2x^2+\tfrac x 2)\operatorname d x}{\int_0^{1/2}(2z+\tfrac 1 2)\operatorname d z}
\end{align}$$
Take it from here.
