What does it mean to integrate a Brownian motion with respect to time?

It might be easier to think about if we abstract a little. Brownian motion is just a continuous, time dependent random variable determined by some probability space $(X,\mathcal F, P)$. Let us view it as a function of two variables, $f(x,t)$. If $x$ is fixed, we have a continuous function, $f_x(t)=f(x,t)$, and we can compute it's integral $F(x,t)=F_x(t)=\int_0^t f_x(s) ds$, where we have just taken the usual integral of a continuous function (Riemann, Lebesgue, whatever, all the definitions agree for nice continuous functions). Thus, we have a transformation of the space of time dependent continuous random variables on $X$ (i.e., functions on $X\times \mathbb R$). The fact that we are working in particular with brownian motion doesn't enter into things.

A more complicated question is what it means to integrate a function or a random variable with respect to Brownian motion. This gets you to the Ito integral (and other similar variants) which are more subtle.

Since Brownian Motion has continuous paths with probability 1 thus$\int_{0}^{t} W_s ds$ is well define. Now set $f(x)=x^3$. By application of Ito's lemma, we have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$ $$W^3(t)=3\int_{0}^{t}W^2_s\,dW_s+3\int_{0}^{t}W_sds$$ therefore $$\int_{0}^{t}W_sds=\frac{1}{3}W^3(t)-\int_{0}^{t}W^2_s\,dW_s$$

Note

The Itô integral can be defined in a manner similar to the Riemann–Stieltjes integral, that is as a limit in probability of Riemann sums; such a limit does not necessarily exist pathwise. Suppose that $W_t$ is a Wiener process and that $X_t$ is a right-continuous (cadlag), adapted and locally bounded process if $I=\{t_0,t_1,\cdots,t_n\}$ is a sequence of partitions of $[0,t]$ with mesh going to zero, then the Itô integral of $X_t$ with respect to $W_t$ up to time t is a random variable $$\int_{0}^{t}X_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{X({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ Set $X_s=W^2_s$, we have $$\int_{0}^{t}W^2_sdW_s=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{W^2({{t}_{i-1}})(W({{t}_{i}})-W({{t}_{i-1}})})$$ Edit \begin{eqnarray} \int_{0}^t W_s \,ds & = & \int_0^t \int_0^t \mathbf{1}_{[0,s]} (u) \,dW_u \,ds \\ & = & \int_0^t \int_0^t \mathbf{1}_{[0,s]} (u) \,ds \,dW_u\\ & = & \int_0^t (t-u) \,dW_u\\ \end{eqnarray} Indeed $$\int_{0}^{t}{{{W}_{s}}}ds\sim N\left( 0\,,\,\frac{1}{3}{{t}^{3}} \right)$$