# For the Brownian motion integrate

I want to calculate $$\operatorname{E} \left[ \int_0^1{W(t)dt \cdot \int_0^1{t^2W(t)dt}} \right].$$

I discovered that the first integral is $$\operatorname{N}(0, \frac{1}{3})$$ but I don't know how to get the other one and the full answer of their multiplied expectation.

Note that \begin{align*} E\left(\int_0^1 W_t\, dt \int_0^1 t^2W_t\, dt \right) &= E\left(\int_0^1\!\!\!\int_0^1 s^2 W_s W_t\, dsdt \right)\\ &=\int_0^1\!\!\!\int_0^1 s^2 E(W_s W_t)\, dsdt\\ &=\int_0^1\!\!\!\int_0^1 s^2 (s\wedge t)\, dsdt\\ &=\int_0^1 s^2\,ds \int_0^1 s\wedge t\, dt\\ &=\int_0^1 s^2\,ds \left(\int_0^s t\,dt + \int_s^1 s\,dt\right). \end{align*} The remaining is now straightforward.