$X\sim Uni[-1, 1]$, $y = x^2$. Find $Cov(x, y)$
$f_X(x) = \frac{1}{R-L} = \frac{1}{2}$
$Cov(x, y) = E(XY)$ <- Would this formula be right?
The solution does it like this: $Cov(x, y) = E(XY) - E(X)E(Y) = E(XY)$.
Is it $E(XY)$ or $E(XY) - E(X)E(Y)$?
No clue how to work with $Y$, by definition this would be
$$E(XY) = \int_{-1}^{1}xy \frac{1}{2}dx = \int_{-1}^{1}\frac{1}{2}x^3dx = \left[ \frac{x^4}{8}\right]_{-1}^{1} = 0$$
Generally, $\mathsf {Cov}(X,Y)= \mathsf E(XY)-\mathsf E(X)\mathsf E(Y)$
However here, $\mathsf E(X) =0$, so that reduces to $\mathsf {Cov}(X,Y)=\mathsf E(XY)$.
$$\mathsf {E}(X)=\int_{-1}^1 \frac{x}2\mathrm d x=0$$
And yes, $\mathsf E(XY)=\mathsf E(X^3)$ which equals $0$ also.
$$\therefore~~\mathsf {Cov}(X,Y)=\int_{-1}^1 \frac{x^3}2\mathrm d x=0$$
PS: If you actually did need to find it, use $\mathsf E(Y)=\mathsf E(X^2)$ and the above method. It equals $\tfrac 13$ .