It is possible to integrate by parts in $\int_0^T B(t) dt$ and obtain

$B(t) (T-t)|_{t=0}^{t=T} - \int_0^T (T-t) dB(t) \overset{d}{=} \int_0^T (T-t) dB(t)$

The Wiener integral on the right has a normal distribution with mean $0$ and variance $\int_0^T (T-t)^2 dt = T^3/3$.

Edit: Sorry, I used $B$ instead of $W$ to denote Brownian motion.

It is possible to integrate by parts in $\int_0^T B(t) dt$ and obtain

$-B(t) (T-t)|_{t=0}^{t=T} + \int_0^T (T-t) dB(t) \overset{d}{=} \int_0^T (T-t) dB(t)$

The Wiener integral on the right has a normal distribution with mean $0$ and variance $\int_0^T (T-t)^2 dt = T^3/3$.

Edit: Sorry, I used $B$ instead of $W$ to denote Brownian motion.