I am confused about something. Let's say $Y=1-X$, $X\sim F_X(x)$ over $(0,1)$ Then $F_Y(y) = P[1-X < y] = P[X> 1-Y] = 1-F_X(1-y)$ Similarly, though theories of expectations, $\mathbb{E}(1-X) = \mathbb{E}(1)-\mathbb{E}(X) = 1 -\mathbb{E}(X)$ Unsing Jensen's inequality also gives this relationship.
However, if$F_X(x) = x^2$ (or I think anything at all over $(0,1)$) then $\int_0^1(F_X(x))\,dx = \int_0^1(F_X(1-x))\,dx$ and $\mathbb{E}(1-X) = 1-\int_0^1(1-x)^2 = 1-\int_0^1(x)^2 = \mathbb{E}(X)$
So now I have that $\mathbb{E}(1-X)= 1 -\mathbb{E}(X)$ and $\mathbb{E}(1-X)= \mathbb{E}(X)$ And since $\mathbb{E}(X)$ is not $\frac{1}{2}$ these can't both be true. What am I doing wrong?
Here is an example that will erhaps this will make my dilemma more clear.
LOGIC 1:
$X\sim F_X(x) = x^2$ over $(0,1)$
$Y = g(X)$, $g(x) = 1-x$, $g^{-1}(x) = 1-x$
$Y \sim F_Y(y) = F_X(g^{-1}(y)) = (1-y)^2$
$\mathbb{E}X = \int_0^1 1- x^2 \,dx = \frac{2}{3}$
$\mathbb{E}Y = \int_0^1 1- (1-y)^2 \,dy =\int_0^1 1- (1-2y+y^2) \,dy = \frac{2}{3}$
Therefore, $\mathbb{E}X=\mathbb{E}Y = \frac{2}{3}$
LOGIC 2:
$\mathbb{E}(Y) = \mathbb{E}(1-X) = \mathbb{E}(1) - \mathbb{E}(X) = 1- \mathbb{E}X = \frac{1}{3}$
There are several different paths to get to both results. Yet, $\mathbb{E}(Y)$ can't equal both $\frac{1}{3}$ and $\frac{2}{3}$
So, somewhere there must be a flaw in reasoning. Help me find it?