Distribution of $\int_{t}^{T} W(s)ds$ where $W$ is Brownian motion

Question

What is the distribution of $\int_{t}^{T} W(s)ds$ given that $W$ is a Brownian motion?

So far, I have the following, $$\int_{t}^{T} W(s)ds = (T-t)W(t) + \int_{t}^{T} (T-s)dW(s)$$ Also, $$\int_{0}^{T} (T-s)dW(s) \sim N(0,\frac{T^{3}}{3})$$

Answer 1

By definition $X=\displaystyle\int_t^TW(s)\mathrm ds$ is Gaussian. Every $W(s)$ is centered hence $X$ is centered. To complete the characterization of the distribution of $X$, it suffices to compute its variance, which is $$ E[X^2]=\int_t^T\int_t^TE[W(s)W(r)]\mathrm dr\mathrm ds=\int_t^T\int_t^T\text{______}=\text{___}.$$