Integral of a Gaussian process has Gaussian Distribution (1) How can we prove that the integral i.e. $\int_{a}^{b} X(t) dt$  (or any linear functional) of a Gaussian process $X(t)$ has Gaussian distribution?
(2) And how can we find that distribution in the case of integral?
Appreciate any insight, thanks.
Edit: (per comment)
$$\lim_{\Delta t \to 0}\sum_{a}^{b}X(t)\Delta t$$
 A: As @Did wrote in his comment we can derive it using the Riemann sum, here is a much longer derivation:
Let $X(t)$ a Gaussian Process:
$$X(t) = GP(m(t), k(t,t'))$$
Then, writing the integral as a Riemann sum:
$$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}\sum_{a}^{b}X(t)\Delta t$$
By the affine property of Multivariate Gaussian random variables we have that:
$$X = \mathcal{N}(\mu, \Sigma) \implies AX = \mathcal{N}(A\mu, A\Sigma A')$$
Hence,
$$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}\sum_{a}^{b}GP(\Delta t \  m(t), \Delta t\ k(t,t')\ \Delta t')$$
Also, we know that the sum of two Gaussians random variables is Gaussian:
If $X \sim \mathcal{N}(\mu_X, \sigma_X^2), Y \sim \mathcal{N}(\mu_Y, \sigma_Y^2)$ and $Z = X + Y$,
then $Z \sim \mathcal{N}(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$.
Hence, we obtain:
$$\int_{a}^{b} X(t) dt = \lim_{\Delta t \to 0}GP(\sum_{a}^{b}\Delta t \ m(t), \sum_{a}^{b}\sum_{a}^{b}\Delta t\ k(t,t')\ \Delta t')$$
Which can also be written as,
$$\int_{a}^{b} X(t) dt = GP(\int_{a}^{b}m(t)\ dt, \int_{a}^{b}\int_{a}^{b}k(t,t')\ dtdt')$$
