Expected value of $Ye^X$ where $X \sim U(0,1)$ and $Y \sim U(0,1)$ I am trying to find the expected value of $Z$ where $Z = Y\cdot e^X$ where $Y \sim U(0,1)$ and $X \sim U(0,1)$.
My attempt so far:
$$F_Z(z) = P(Ye^X \le z) = \int \int_{Ye^X \le z} f(x,y)\, dxdy$$
Where $f_{X,Y}(x,y) = f_y\cdot f_{e^x}$
$$f_Y(y) = \frac{1}{1-0}\,, \quad  y \in (0,1)$$
I am stuck trying to find $f_{e^X}$ but I cannot remember how to find that pdf.
 A: If the problem does not state explicitly that $X$ and $Y$ are independent, then it doesn't have a solution, because the marginal distributions of $X$ and $Y$ do not determine their joint distribution.
Supposing that $X$ and $Y$ are independent, then $e^X$ and $Y$ are also independent. Proof:
$$
\begin{eqnarray}
  P\left\{e^X \in A, Y\in B \right\} &=& P\left\{X \in \exp^{-1}(A), Y\in B \right\} \\
 &=& P\left\{X \in \exp^{-1}(A)\right\}P\left\{Y\in B \right\} \quad \textrm{[independence of $X$ and $Y$]} \\
  &=& P\left\{e^X \in A\right\}P\left\{Y\in B \right\} \, ,
\end{eqnarray}
$$
in which $A$ and $B$ are Borel sets, and $\exp^{-1}(A)$ is the inverse image of the set $A$ under $\exp$. Note that $\exp^{-1}(A)$ is also a Borel set because $\exp:\mathbb{R}\to\mathbb{R}$ is measurable.
But the expectation of a product of independent random variables is the product of their expectations. Therefore,
$$
 \mathrm{E}\left[Ye^X\right]=\mathrm{E}[Y]\cdot\mathrm{E}\left[e^X\right] \, .
$$
To compute $\mathrm{E}\left[e^X\right]$ you don't need to determine the distribution of $Z=e^X$. It follows from a theorem, known folkloricaly as the Law of the Unconscious Statistician,  that
$$
  \mathrm{E}\left[e^X\right] = \int_{-\infty}^\infty e^x f_X(x)dx \, ,
$$
and this integral is easy to compute.
A: There is definitely an easier approach to this problem (hints were given in the comments), but since you asked about a specific step, I'll go from there.
You want to compute the pdf $f_{e^X}(x)$.  Let's start with:
$$ F_{e^X}(x) = P(e^X < x)
              = P(X < \log x)
              = (F_X \circ \log)(x)
$$
Recall that we can compute $f_{e^X}$ by differentiating $F_{e^X}$.  In this case, you can use the chain rule.
This kind of transformation generalizes to the multivariate case.
