How to solve for the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$? I would like to find the expectation of the Ito Integral: $\int_0^4 B_t^2 dB_t$. My strategy is to use Ito's general formula with:

$$
f(t, B_t) = f(0,0) + \int_0^t \frac{df}{dx}(s, B_s) dB_s + \int_0^t \frac{df}{dt}(s, B_s) ds + \frac{1}{2}\int_0^t \frac{d^2f}{dx^2}(s, B_s) ds
$$

Then, my candidate function is $f(t, B_t) = \frac{B_t^3}{3}$. However, the problem I am having is in the derivative $\frac{df}{dt}(s, B_s)$. Since I want to have my integration go up to $4$, then I necessarily have:

$$
\int_0^4 \frac{df}{d4}(s, B_s) ds = \int_0^4 \frac{\frac{B_t^3}{3}}{d4} ds
$$ 

I know that there is probably a bad/stupid mistake here, and that I should just take the derivative with respect to $s$, but I cannot figure out why exactly we are having the $dt$ MATCH the upper limit of the integral, $t$. Can anyone shed some light here? Thanks!
Let $f(x) = \frac{x^3}{3}$ and for reference, note that $f'(x) = x^2$ and $f''(x) = 2x$. It\^o's formula gives us that \begin{eqnarray*}
f(B_t) - f(0) &amp;=&amp; \int_0^t f'(B_s) dB_s + \frac{1}{2} \int_0^t f''(B_s) ds. 
\end{eqnarray*} Therefore, we see that \begin{eqnarray*}
\frac{1}{3} B_t^3 &amp;=&amp; \int_0^t B_s^2 dB_s + \int_0^t B_s ds \\
\therefore \ \mathbb{E} \left( \frac{1}{3} B_t^3 \right) &amp;=&amp; \mathbb{E} \left( \int_0^t B_s^2 dB_s + \int_0^t B_s ds \right) \\
\therefore \ \frac{1}{3} \mathbb{E} (B_t^3) &amp;=&amp; \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \ \ \ \text{(linearity of expectation)}, \\
0 &amp;=&amp; \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) + \mathbb{E} \left( \int_0^t B_s ds \right) \\
\therefore \ \mathbb{E} \left( \int_0^t B_s^2 dB_s \right) &amp;=&amp; - \mathbb{E} \left( \int_0^t B_s ds \right) =  - \int_0^t \mathbb{E}(B_s) ds \ \ \ \text{(Fubini)} \\
&amp;=&amp; 0, 
\end{eqnarray*} where the last line follows from the fact that the Brownian motion is normally distributed with mean zero.
Choose $$f(t, B_t) = \frac{B_t^3}{3}$$

so that we have

$$
\frac{B_t^3}{3} = \int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds
$$

$$
\to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds]
$$

$$
\to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] + \frac{1}{2}\int_0^t 2E[B_s] ds
$$

$$
\to E[\frac{B_t^3}{3}] - \frac{1}{2}\int_0^t 2E[B_s] ds = E[\int_0^t B_s^2 dB_s]
$$

$$
\to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s]
$$

$$
\to \frac{E[B_t^3]}{3} \stackrel{?}{=} \frac{0}{3} = E[\int_0^t B_s^2 dB_s]
$$