Variance of $\int_0^1X(t)dt$

Question

Let $X(t)$ be a stationary random variable with expected value $E[X(t)] = m$ and covariance function $r_X(\tau) = 2e^{-|\tau|}$.

I'm asked to calculate the variance of $\int_0^1X(t)dt$, $$V[\int_0^1X(t)dt].$$ I've tried using the formula $$C[X,Y] = E[XY] - E[X]E[Y]$$ but I can't figure out $$E[\int_0^1X(t)dt\cdot\int_0^1X(t)dt].$$ How do I do it? Note; all the information given may not be necessary.

The answer is supposedly $4e^{-1}.$

Answer 1

As kimchi lover says, \begin{align} E\int_0^1\int_0^1(X_t-m)(X_s-m)dsdt & = 2E\int_0^1\int_0^t(X_t-m)(X_s-m)dsdt\\ & = 2\int_0^1\int_0^t2e^{-(t-s)}dsdt\\ & = 4e^{-1}. \end{align}

Answer 2

Variance$=E(\int_0^1(X(t)-m)dt)^2)={\int_0^1\int_0^1E(X(t)-m)(X(s)-m))dtds}$

$=\int_0^1\int_0^12e^{-|t-s|}dtds=I$. This integral is a bit messy. $I=2\int_0^1(e^{-s}\int_0^s e^tdt+e^s\int_s^1e^{-t}dt)ds=\frac{4}{e}$.