How to calculate conditional expectation $\mathbb{E}[g(x)\ | \ x\geq Q]$ for $x\sim \exp(1)$ Given that $X$ is exponentially distributed continuous random variable $X\sim \exp(1)$ and $g(x)$ is as below. How can I find the Expectectaion of $g(x)$ for the condition that $x\geq Q$, i.e. $\mathbb{E}[g(x)\ | \   x\geq Q]$.
$$g(x) = \frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big)$$
I suppose I should start by considering an event $\Phi$ such that $\Phi = \mathbb{P}[x \geq Q]$. However, I am stuck about how to go around this condition.
All constants are positive real values.

EDIT: Is the following solution correct for the above question? If it is OK, then I have found the solution. But I will really appreciate if someone can let me know if the following method is correct.
$$\mathbb{E}[g[x]] = \int_Q^\infty g(x). f(x) dx$$
$$\mathbb{E}[g[x]]= \int_Q^\infty\frac{A}{\exp(-bQ+c)}\Big(\frac{1 + \exp(-bQ+c)}{1 + \exp(-dx+c)}-1\Big) e^{-x} dx$$
 A: I think the solution you gave is off by a normalization constant.
To see this, consider $g(x) = 1$. Then your formula gives $\mathbb{E}[g(x)\ |\ x\geq Q] = \int_Q^\infty f(x) dx \leq 1$, while the correct answer should be identically one: the expectation of a constant is always just that constant.
The correct expression can be found by augmenting the "problem space" $X$ to $(X,U)$, where $X \in \mathbb{R}^+$ is the original random variable of your question and $U \in \{0,1\}$ is a discrete binary variable which is $1$ if $X \geq Q$ and $0$ otherwise. Then we can write the problem in a more manageable form:
$$
\begin{align}
\mathbb{E}[g(x)\ |\ x\geq Q] &\equiv \int dx \ g(x) \ p(x \ | \ x\geq Q) \\
&= \int dx \ g(x) \ p(X = x \ | \ U = 1).
\end{align} \tag{1} \label{1}
$$
Now we can use the product and sum rules:
$$
p(X = x \ | \ U = 1) = \frac{p(U = 1 \ | \ X = x) \ p(X = x)}{\int dy \ p(U = 1 \ | \ X = y) \ p(X = y)}. \tag{2} \label{2}
$$
With the conditional pmf
$$p(U = 1 \ | \ X = x) = \begin{cases}
0 & x < Q \\
1 & x \geq Q
\end{cases} \tag{3} \label{3}
$$
and setting $p(X = x) = f(x)$, we get after substituting \eqref{3} in \eqref{2} and then \eqref{2} into \eqref{1}:
$$
\mathbb{E}[g(x)\ |\ x\geq Q] = \frac{\int_Q^\infty dx \ g(x) \ f(x)}{\int_Q^\infty dx \ f(x)}.
$$
