Here is the question find $E(X\mid Y=y)$ when $f(x,y) = 6(1-y)$ on the region $0 \le x \le y \le 1$ and $0$ elsewhere.
In setting this up I first found $f_y = \int_0^y 6-6y \ dx$ which yields $f_y = 6(y-y^2)$
From here, I used the formula for finding the conditional expected value hence: $\int_0^1\int_x^1\frac{6y(1-y)}{6(y-y^2)}\ dy\ dx$ which reduces to $\int_0^1\int_x^1 \ dy \ dx = \frac{1}{2}$
The answer in the back of my book says that it should be $\frac{y}{2}$
What did I do wrong?