Q) Let $W = \int_{0}^{t}B_sds$, $B$ is a Brownian motion. Find $EW$ and $EW^2$.
My attempt: $B_s \sim N(0,s) $
$$W = \int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds , EW = \int_{-\infty}^{\infty}x(\int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds)dx$$
I am not sure how to find $\int_{0}^{t}\frac{1}{\sqrt{2\pi s}}e^{-\frac{x^2}{2s}}ds$? Integration by reduction makes sense in the case of $\int x^ne^{ax}dx$ if n is a positive integer, where in if $I_n = \int x^ne^{ax}dx$, then $I_n = \frac{1}{a}(e^{ax}x^{n+1}-aI_{n-1})$, $I_0$ can be computed and all $I_n$ recursively.
Thanks and appreciate a hint.
This is relatively straightforward if you can use Ito's Lemma and Ito's Isometry.
If you are allowed to use those result than $\mathbb E W$ and $\mathbb E W^2$ can be computed in the following way. Start by applying Ito's Lemma to $t B_t$ to obtain
$$ d(tB_t)=B_tdt+tdB_t$$
From this we obtain that
$$\int_0^t B_sds = t B_t - \int_0^t s dB_s$$ It is immediate form the above expression that $\mathbb E W =0$. To compute $\mathbb E W^2$, note that the expression for $\int_0^t B_sds$ derived above implies that $$ \int_0^t B_sds = t \int_0^t \left(1-\frac{s}{t}\right)dB_s$$ Therefore, $$ \mathbb E W^2 = t^2\mathbb E\left[\left(\int_0^t \left(1-\frac{s}{t}\right)dB_s\right)^2\right] = t^2 \int_0^t \left(1-\frac{s}{t}\right)^2ds$$ by Ito Isometry. Computing this integral, we note that $\mathbb E W^2=t^3/3$.