Compute $\int_1^2 B_t \; dB_t$

Question

I have to compute the following Ito integral: $$\int_1^2 B_t \; dB_t$$ where $(B_t)_{t \geq 0}$ is the 1-dimensional Brownian Motion.

In the definition of Ito integral, the integral is taken from $0$ to some $T$, i.e we have $\int_0^T$. But I have to calculate $\int_1^2 B_t \; dB_t$.

Can someone, please, explain me/give me a hint on how to calculate this integral?

Thank you!

Answer 1

let $f(x)=x^2$ by application of Ito's lemma, we have $$f(B_t)=f(B_s)+\int_{s}^{t}f'(B_u)\,dB_u+\frac{1}{2}\int_{s}^{t}f''(B_u)du$$ as a result $$B_t^2=B_s^2+2\int_{s}^{t}B_u\,dB_u+\int_{s}^{t}du$$ let $t=2$ and $s=1$, thus $$B_t^2=B_s^2+2\int_{1}^{2}B_u\,dB_u+\int_{1}^{2}du$$ in the other words $$\int_{1}^{2}B_u\,dB_u=\frac{B^2(2)-B^2(1)-1}{2}$$