where's the error of this calculation?

Question

Let $X=2e^{-2\theta}$, ask to calculate $E[X^2]$. Here's the right way to calculate it $$E[X^2]=var[X]+(E[X])^2={1\over 2^2} + ({1\over 2})^2={1\over 2}$$ but if I do it in another way $$E[X^2]=E[(2e^{-2\theta})^2] = E[4e^{-4\theta}]= {1\over 4}$$ Then it ends up $1\over 4$, what's wrong with the second calculation?

Answer 1

\begin{equation} E[X^2] = \int_0^{\infty} x^2 2e^{-2x} dx = \frac{1}{2} \end{equation} $X=2e^{-2x}$ is confusing, since $X$ is a function that has the distribution $2e^-2x$.