Using Ito’s formula, write an expression for $\int_0^1(B(s))^2dB(s)$
Not sure exactly if I did this right. Was hoping for feedback. I let $f(x)=x^3$.
Then by definition, Itos formula states that
$f(B(b))-f(B(0)) = \int_0^bf'(B(s))dB(s) + \frac{1}{2}\int_0^1f''(B(s))ds$
I then said that:
$(B(1))^3-(B(0))^3=\int_0^13(B(s))^2dB(s)+\frac{1}{2}\int_0^16(B(s))ds$
$B(1))^3 - 0 - 3\int_0^1B(s)ds = 3\int_0^1(B(s))^2dB(s)$
so then
$ \int_0^1(B(s))^2dB(s)=\frac{(B(1))^3}{3}-\int_0^1B(s)ds$
Did I do this right? If so can the right hand side be reduced any more? Thanks
