I'd like to see what $E(X|X>a)$ means when $X = \text{Exp}(1)$
First interpretation would be
$$ \dfrac{1}{\int_a^{\infty} \text{pdf}(x)\mathrm{d}x} \int_a^{\infty} x\cdot\text{pdf}(x) \mathrm{d}x$$
Here we are using the probability of in our time frame, thinking anything can happen but just want to sum up the $x\cdot p(x)$ but only over $X>a$
Another interpretation would be
$$\int_a^{\infty} x\cdot \text{pdf}(x-a) \mathrm{d}x$$
Where you apply $\begin{cases}\Pr(X<t|X>a) =0&\text{if }t<a\\ \Pr(X-a<t)&\text{ if }t>a \end{cases}$
Which can also be expressed by
$\mathbb{E}(X|x>a) = \mathbb{E}(X-a+a|x>a) = a + \mathbb{E}(X-a|x>a) = a + \mathbb{E}(X) = a+1 $
then this means, you are thinking of the events where $X>a$ only so there's no probability of $X<a$ happening.
I wonder if the first interpretation makes sense as well?
Yet another way would be $ E(X) = \int_0^a xe^{-x} dx + {\int_a^\infty xe^{-x}dx }$ So $ {\int_a^\infty xe^{-x}dx } = E(X) - { \int_0^a xe^{-x}dx } $ but I guess that's not what people mean
The first works as a general statement of the conditional expectation (and can also be described as "thinking of the events where $X>a$ only so there's no probability of $X≤a$ happening") while the second uses the memoryless property of the exponential distribution. In this case you get $a+1$ for both methods.
For the first you get for $a \ge 0$: $$\frac{\int_a^\infty xe^{-x}\,dx }{\int_a^\infty e^{-x}\,dx} = \frac{(a+1)e^{-a} }{ e^{-a}} = a+1$$