How can I solve this expected value? Good evening, how can I solve this expected value?
$$
E \Bigl[ B_1 \int_0^{x} B_u du\ \Bigr]
$$
where $B_t$ is a standard Brownian Motion and x > 0.
 A: Integration by parts justifies
$$
\int_{0}^{x}B_u \mathrm du = xB_x - \int_{0}^{x}u \mathrm dB_u\,.
$$
Therefore,
$$
\begin{array}{rcl}
\displaystyle \mathbb{E}\left[B_1 \int_{0}^{x}B_u \mathrm du\right] &{}={}&\displaystyle  \mathbb{E}\Big[xB_1 B_x \Big] - \mathbb{E}\left[B_1 \int_{0}^{x}u ~\mathrm dB_u \right] \\
&{}={}&\displaystyle   x\mathbb{E}\Big[B_1 B_x \Big] - \mathbb{E}\left[ \int_{0}^{1} 1 ~\mathrm dB_u \int_{0}^{x}u ~\mathrm dB_u \right] \\&{}={}&\displaystyle  x\mathbb{E}\Big[B^2_{1\wedge x} \Big] - \mathbb{E}\left[ \int_{0}^{1\wedge x} 1 ~\mathrm dB_u \int_{0}^{1\wedge x}u ~\mathrm dB_u \right] \\ &{}={}&\displaystyle   x\cdot\left(1\wedge x\right) - \int^{1\wedge x}_{0}u~\mathrm du \\ &{}={}&\displaystyle   \left(x - \frac{1}{2}\left(1\wedge x\right)\right)\cdot\left(1\wedge x\right)\,. 
\end{array}
$$
Above, use was made of the "independent increments", "variance" and "zero expectation" properties of Brownian motion. Also, Ito's Isometry (see this answer) justifies 
$$
\mathbb{E}\left[ \int_{0}^{1\wedge x} 1 ~\mathrm dB_u \int_{0}^{1\wedge x}u ~\mathrm dB_u \right] = \int^{1\wedge x}_{0}u~\mathrm du\,.
$$
We have just shown,

$$
\mathbb{E}\left[B_1 \int_{0}^{x}B_u \mathrm du\right] = \left(x - \frac{1}{2}\left(1\wedge x\right)\right)\cdot\left(1\wedge x\right)\,.
$$

A: This article might be useful. You evaluate the integral as follows: $\int_0^x B_u du = (B_0 + B_h + B_{2h} + ... + B_x)h$ with a stepwidth $h$. Then you can use the fact for Standard Brownian motion $E(B_i B_j) = \delta_{ij}$ for some numbers $i,j$.
