Consider a standard Brownian Motion $X_t$ and continuous random variable $Y_t$, where $Y_t$ is defined as $$ Y_t = \int_0^t X_t \, dt $$
My goal is to compute the variance of $Y_t$. I'd like to say that $\operatorname{Var}(Y_t)=\int_0^t \operatorname{Var}(X_t) \, dt$, but I'm not sure how I'd establish that the variance of an integral is the integral of the variance, or if that's even true. Is it true, and if so, how can I show it?