Limits of integration of manipulated probability distribution function I have to compute $\mathbb E\left[e^X \right]$, where $X$ follows a uniform distribution on $[0,1]$.
I have started by computing the probability density function of $Y = e^X$, by doing the following:
$$
F_Y(x) = \mathbb P(Y \leq x) = \mathbb P(e^X\leq x) = \mathbb P(X\leq \log(x)) = \int_0^{\log(x)}f(y)\,dy = \log(x)
$$
where $f(x)$ is the probability density function of the uniform random variable.
Now, in the final step, when I have to compute the expectation $\mathbb E[Y]$, I am unsure which limits of integration to choose. 
I have the feeling it is no longer $0$ and $1$, and that is the question precisely. How do the limits of integration change, when changing the probability distribution function? (for example, in this case, the random variable has exponentiated).

This question is easier to do directly so I'll do both (btw. this is just the moment generating function of the random variable with $t=1$ http://en.wikipedia.org/wiki/Moment-generating_function).
$X$ can take values in $[0,1]$ so $e^X$ can take values in $[e^0,e^1]=[1,e]$. Let's now use your work and see what we get,
\begin{align*}
F_Y(x)=log(x) \Rightarrow f_Y(x)=\frac{1}{x}\\
E[Y]=\int_1^e x\frac{1}{x}dx=x|_1^e=e-1
\end{align*} 
Doing the problem directly now (as @V.C. just did),
\begin{align*}
E[Y]=E[e^X]=\int_0^1e^xdx=e^x|^1_0=e-1
\end{align*}
