I was studying an example given in Variance of a time integral with respect to a Brownian Motion function
There we need to calculate the Variance of $I_t = \int\limits_{0}^{t} f \left(s \right) W_s ds$.
Then, it said that $E \left[ I_t^2 \right] = \int\limits_{0}^{t}\int\limits_{0}^{t} f \left(s \right) f \left(u \right) \min \left(s,u\right) dsdu$
Can someone please help to understand the detailed calculation on how it can be arrived?
It is also stated that $I_t$ follows a Normal distribution. What is the reason for that?