Find the pdf of Y = 1/X and compute E(Y) Let the random variable have a pdf $f_X(x) = 2x$ when $0<x<1$ and $0$ otherwise. 
Let the random variable $Y=1/X$. Find the pdf of $Y$ and use it to calculate $E(Y)$. 
The pdf of $Y$ is $f_Y(y) = \frac{2}{y^3}$. What is the support for $f_Y(y)$? 
 A: The first thing to do in such a problem is to figure out the
support of the random variable $Y$, which is $(1, \infty).$
To find it, try several values of $x$ in $(0, 1)$ and see what 
happens. You were right to ask about this crucial issue.
Your text probably shows transformation methods based on the CDF
and on the PDF. An abbreviated version of the former (with gaps for you to complete) is as follows:
$$ F_Y(t) = P(Y \le t) = P(1/X \le t) = P(X \ge 1/t)\\ = 1 - P(X \le 1/t) = 1 - 1/t^2,$$
for $t$ in the support of $Y$. 
Then, by differentiation, the 
PDF of $Y$ is $f_Y(t) = 2t^{-3},$ as you say.
Finally, $E(Y) = \int_1^\infty 2t^{-3} \, dt = 2.$
[Some of this is in your Question, and some is in @drhab's
succinct Answer (+1)]
If you recognize that $X \sim Beta(2, 1),$ then it is easy to do
a quick simulation in R to check this answer (correct to
two or three decimal places based on a million simulated
values of $Y$):
 x = rbeta(10^6, 2, 1);  y = 1/x;  mean(y)
 ## 2.001900  # aprx of E(Y)

Below is a histogram of 10,000 simulated values of $Y$
along with a plot of the PDF of $Y$. This distribution
is extremely right-skewed, with a long tail to the right.
The scale of the histogram extends out to about 60 in recognition of
occasional scattered points too sparsely spread to
make histogram bars of noticeable height. (The biggest
of the 10,000 values plotted in this histogram was at 57.44494.)

A: Based on the data we find: $$F_{X}\left(x\right)=\begin{cases}
0 & \text{if }x\leq0\\
x^{2} & \text{if }0<x<1\\
1 & \text{if }x\geq1\end{cases}$$
Based on that we find: $$F_{Y}\left(y\right)=\begin{cases}
0 & \text{if }y\leq1\\
1-y^{-2} & \text{if }y>1\end{cases}$$
Then $f_Y$ prescribed by: $$y\mapsto\begin{cases}
0 & \text{if }y\leq1\\
2y^{-3} & \text{if }y>1\end{cases}$$serves as PDF (it is the derivative of the CDF).
