Expectation of the integral of Wiener process?

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Background

Consider the Wiener process: \begin{equation} W(t)=\int_0^t d W=\int_0^t \xi\left(t^{\prime}\right) d t^{\prime} \end{equation} where $\xi(t)$ is white Gaussian noise: $\langle \xi(t)\rangle = 0$ and $\langle \xi(t)\xi(t')\rangle = \delta(t-t')$.

Consider now the following stochastic integral:

$$ I=\int_0^T f\left(t^{\prime}\right) d W=\int_0^t f\left(t^{\prime}\right) \xi\left(t^{\prime}\right) d t^{\prime}$$

Where $f(t)$ is some function of $t$. If $f(t)$ is the Wiener process itself, we can show that $\mathbb{E}[ I]$ = 0 under the Ito interpretation, and $T/2$ in the Stratonovich interpretation.

Question

What is $\mathbb{E}[I]$ if I choose $f$ to be the integral of the Wiener process? I.e. let $$f(t)=\int_0^t W(t') d W$$

Edited Changed $\langle .\rangle$ to $\mathbb{E}[.]$.