I am trying to evaluate
$$W=\int_0^t B_s ds$$
where $B_s$ is a one-dimensional Brownian motion, with $B_0=0$. My attempt is as following:
As $s\mapsto B_s$ is continuous with probability one, it is Riemann integrable, and hence \begin{align*} \int_0^t B_sds&=\lim_{n\to\infty}\sum_{k=1}^n B\left(\frac{tk}{n}\right)\frac{1}{n}\\ &=\lim_{n\to\infty}\sum_{k=1}^n \sqrt{\frac{k}{n}}B_t\frac{1}{n}\\ &=B_t\int_0^1\sqrt{x}dx=\frac{2}{3}B_t. \end{align*}
I used the property that $B_{st}$ and $t^{1/2}B_s$ have same law. This says that $EW=0$ and $EW^2=\dfrac{4}{9}t$.
However, some StackExchange answer says that its variance is $\dfrac{t^3}{3}$. The method in the link seems valid, but I cannot find what is wrong with my answer.
Can someone explain why my method is wrong?
P.S. This is an exercise of Durrett 7.1.3. In this chapter, it is not introduced neither Itô calculus, nor martingale property of BM.
The said exercise does not ask you to evaluate the integral, merely to find its distribution. As Chris has rightfully pointed out in their comment, you can't say that $B_{at} = B_t\sqrt a$. You should not forget that we use to omit $\omega$ as an argument when dealing with random variables. Imagine you have $X,Y$ iid random variables, so that their distribution are indeed the same, but definitely for quite some $\omega$ we will have $X(\omega)\neq Y(\omega)$. Same story with $B_{at}$ and $B_t\sqrt a$.
Overall, I don't think that $\int_0^t B_s\mathrm ds$ has any simpler form. It is definitely not expressible in a form of $G(t, B_t)$ for any $G$. Why? The latter is a Markov process, its future distribution only depends on its current value, it has no memory dependency. Your integral certainly does depend a lot on all values of $B_s$ for $s\in[0,t]$.
Hence, I would suggest you to just find mean and variance of this integral as you're asked in the exercise, and also think about why would it be normally distributed.