Consider the following integral:
$$I = \int_0^L \omega(t) \cdot X(t)\,\mathrm{d}t$$
Where $X(t)$ is a stationary stochastic process with mean $\mu$, variance $\sigma^2$ and correlation function $\rho(\tau)$ where $\tau = |t_1 - t_2|$ and $\omega(x)$ is a deterministic weighting function satisfying:
$$\int_0^L \omega(t)\,\mathrm{d}t = 1$$
Then:
$$\mathbb{E}\left[I\right] = \int_0^L \mathbb{E}\left[\omega(t)\right] \cdot \mathbb{E}\left[X(t)\right]\,\mathrm{d}t$$
Noting that $\mathbb{E}\left[\omega(t)\right] = \frac{1}{L}\int_0^L\omega(t)\,\mathrm{d}t = \frac{1}{L}$ and then $\mathbb{E}[X(t)] = \mu$ then:
\begin{equation} \mathbb{E}\left[I\right] = \frac{\mu}{L}\int_0^L \,\mathrm{d}t = \mu \end{equation}
Now turning our attention to the variance of $I$ which is given by $\mathbb{E}\left[I^2\right] - \left(\mathrm{E}[I]\right)^2$:
$$\mathbb{E}[I^2] = \mathbb{E}\left[\int_0^L \omega(t) \cdot X(t)\,\mathrm{d}t \int_0^L \omega(u) \cdot X(u)\,\mathrm{d}u\right]$$
$$\mathbb{E}[I^2] = \mathbb{E}\left[\int_0^L \int_0^L \omega(t)\,\omega(u)\,X(t)\,X(u)\,\mathrm{d}t\,\mathrm{d}u\right]$$
Can anyone continue this calculation to obtain $\mathbb{E}[I^2]$ in terms of $\sigma$ and $\rho(\tau)$?
Since $\omega$ is a deterministic function, we can bring it outside whenever we take expectation with respect to $\mathbb{E}$. So we have $$\mathbb{E}[I] = \mathbb{E}\Big[\int_0^L \omega(t)X(t)dt \Big] = \int_0^L\omega (t)\mathbb{E}[X(t)]dt $$ $$\mathbb{E}[I]^2 = \Big( \int_0^L\omega (t)\mathbb{E}[X(t)]dt\Big) \Big( \int_0^L\omega (s)\mathbb{E}[X(s)]ds\Big) = \int_0^L \int_0^L \omega(t)\omega(s)\mathbb{E}[X(t)]\mathbb{E}[X(s)]\ dt ds $$ Similarly we have $$ I^2 = \int_0^L \int_0^L \omega(t)\omega(s) X(t)X(s)\ dt ds $$ $$ \mathbb{E}[I^2] = \int_0^L \int_0^L \mathbb{E}[\omega(t)\omega(s) X(t)X(s)]\ dt ds = \int_0^L \int_0^L \omega(t)\omega(s) \mathbb{E}[X(t)X(s)]\ dt ds$$ And so we obtain $$\begin{split} Var(I)=\mathbb{E}[I^2]-\mathbb{E}[I]^2 & = \int_0^L \int_0^L \omega(t)\omega(s) \{ \mathbb{E}[X(t)X(s)]-\mathbb{E}[X(t)]\mathbb{E}[X(s)]\}\ dt ds\\ & = \int_0^L\int_0^L \omega(t)\omega(s)Cov(X(t),X(s))\ dt ds \end{split}$$ In your case, since $Cov(X(t),X(s))=\rho(\vert t-s\vert)$, the expression becomes $$Var(I)= \int_0^L\int_0^L \omega(t)\omega(s)\rho(\vert t-s\vert) dtds$$