I'm studying the use of Brownian motion (aka Wiener Process) in the context of the Doob's optional stopping theorem for continuous cases. I do remember the basic properties of a Brownian motion (stationarity, independence or the increments, the fact that $W(t) \sim N(0, t)$), but I struggle with a "trivial" point on the proof of the fact that $Cov (W(t_1), W(t_2)) = \min(t_1, t_2)$.
By using the different properties I reach the point where I can say that, for $t_1 < t_2$, $Cov (W(t_1), W(t_2)) = E((W(t_1))^2) = t_1$.
Maybe my brain used all its energy but I can't find why $E(W(t_1)^2) = t_1$, knowing that the first moment is $0$.