If $X$ is a random variable and $A$ is an event with $P(A) > 0$, then the conditional expectation of $X$ given $A$ is defined
$$E[X \mid A] = \frac{1}{P(A)} \int_{A} X\ dP.$$
However, if $X$ is continuous and has density $f$, how do I write the above in terms of an integral over $f$, so that I can actually carry out this computation? Would this be equal to
$$\frac{1}{P(A)} \int_B x f(x)\ dx$$
where $B = X(A) = \{ X(\omega) : \omega \in A\}$?
$\int_{A} X\ dP$ cannot be written in terms of $f$ in general. You need the density of $XI_A$. If $A$ is of the form $X^{-1}(E)$ for some Borel set $E$ in $\mathbb R$ the you can write $\int_{A} X\ dP=\int_E xf(x)dx$.