Let $W=\int_0^t B_sds$. Find $EW$ and $EW^2$. What is the distribution of $W$?
From the definition, we have $$ EW=\int_0^t EB_sds=0 $$ and $$ EW^2=\int_0^t EB^2_sds=\int_0^t sds=t^2/2. $$
But how to describe the distribution of $W$? Does it sound like $W\sim N(0, t^2/2)$?
For each $n\ge 1$, let $\Delta s:=t/n$ and for $k=0,\ldots, n-1$ let $s_k:=k\Delta s$. Consider the Riemann sum approximation of $W$: $$ W^{(n)}=\Delta s\sum_{k=0}^{n-1}B(s_k). $$ Since $W^{(n)}\to W$ as $n\to\infty$ and each $W^{(n)}$ is Gaussian, the limit is also Gaussian. Trivially, $\mathsf{E}W^{(n)}=0$, and since $\operatorname{Cov}(B(s_k),B(s_j))=\min\{s_k,s_j\}$, $$ \mathsf{E}\big(W^{(n)}\big)^2=\frac{t^2}{n^2}\left(\frac{tn^2}{3}-\frac{tn}{2}+\frac{t}{6}\right)\to \frac{t^3}{3} $$ as $n\to\infty$. Thus, $W\sim N(0, t^3/3)$.