Uniform Random Variable: Correlation and Independence Let X be a uniform random variable defined on the interval $(0,1)$. If $Y = 6X^2−6X+1$, compute the correlation of X and Y . Are X and Y independent? Are X and Y uncorrelated?
So my work is. 
$F(X) = \int{1} dx = x$
$E(X) = \frac{a+b}{2} = \frac{1}{2}$
$F(Y) = \int 6X^2-6X+1 dx = 2x^3-3x^2+x$
So is $E(XY) = \int\int f(x)f(y)dxdy$?
 A: $X$ is a uniform continuous random variable on the support $(0;1)$.
Some features of this are that it has a cummulative distribution function $F_X(x) = x\;\raise{0.25ex}\chi_{x\in(0;1)}$, a probability mass function of $f_X(x)=1\,\raise{0.25ex}\chi_{x\in(0;1)}$, an expectation $\mathsf E(X)= \frac 1 2$ and variance: $\mathsf E(X^2)-\mathsf E(X)^2 = \frac 1 {12}$
Indeed $\mathsf E(X^k) = \int_0^1 x^k\,f_X(x)\operatorname d x = \frac{1}{k+1}$ for all $\Re (k)>-1$
Use this.  

$$\begin{align}
\mathsf E(X^k) & =\frac{1}{k+1}
\\[2ex]
\mathsf E(XY) & = \mathsf E(6X^3-6X^2+X)
\\[1ex] & = 6\mathsf E(X^3)-6\mathsf E(X^2)+\mathsf E(X)
\\[1ex]
& =
\\[2ex]
\mathsf {Cov}(X,Y) & = \mathsf E(XY)-\mathsf E(X)\mathsf E(Y)
\\[1ex]
& = \mathsf E(6X^3-6X^2+X)-\mathsf E(X)\mathsf E(6X^2-6X+1)
\\[1ex]
& = 6\mathsf E(X^3)-6\mathsf E(X^2)+\mathsf E(X)-\mathsf E(X)\Big(6\mathsf E(X^2)-6\mathsf E(X)+\mathsf E(1)\Big)
\\[1ex] 
& =
\end{align}$$
A: That's too much work, you would rather do
$$E[XY] = E[X(6X^2-6X+1)].$$
Then use the properties of expectation.

For the covariance, I would proceed as follows:
\begin{align*}
\text{Cov}(X,Y) &= \text{Cov}(X,6X^2-6X+1)\\
&=6\text{Cov}(X,X^2)-6\text{Cov}(X,X)+\text{Cov}(X,1)\\
&=6\left[E[X^3]-E[X]E[X^2]\right]-6\text{Var}(X)+0\\
&=6\left[\int_0^1x\cdot x^3\,dx-\frac{1}{2}\left(\frac{1}{12}+\frac{1}{4}\right)\right]-\frac{1}{2}\\
&=6\left[\frac{1}{5}\cdot 1 -\frac{1}{2}\cdot \frac{4}{12}\right] -\frac{1}{2}\\
&=-\frac{3}{10}.
\end{align*}
