From the post Integral of Brownian motion w.r.t. time we have an argument for
$$\int_0^t W_sds \sim N\left(0,\frac{1}{3}t^3\right).$$
However, how does this generalise for the interval $[t;T]$? I.e. what is the distribution of
$$\int_t^T W_sds.$$
I would expect it to be $$\int_t^T W_sds \sim N\left(0,\frac{1}{3}(T-t)^3\right),$$
but I cannot see why.