Compute $\mathbb{E}(\int_{0}^{T} |B_{t}|^{2} dt)$

Question

For $(B_t)_{t\geq0}$ a standard brownian motion and $T$ the first time the standard brownian motion hits $1$ or $-1$ I have to compute $\mathbb{E}(\int_{0}^{T} |B_{t}|^{2} dt)$. My lecturer has given us a hint to first apply Ito's formula to $B_t^4$. There are two things which kind of make sense to me and I'm not sure which is correct if either. First, using the hint and then taking expectation I think we get

$\mathbb{E}(|B_t|^4) = 6\mathbb{E}(\int_{0}^{t} |B_{s}|^{2} ds)$,

then to evaluate this at $t=T$, we get

$\mathbb{E}(|B_T|^4) = 6\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds)$,

but $|B_T| = 1$, so $\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds) = \frac{1}{6}$.

That's the first way that kind of makes sense. The second way is ignoring the hint and using Fubini's theorem to get

$\mathbb{E}(\int_{0}^{T} |B_{s}|^{2} ds) = \int_{0}^{T} \mathbb{E}(|B_{s}|^{2}) ds = \int_{0}^{T} s ds = \frac{T^2}{2}$.

I'm slightly confused as to whether either of these ways is correct and if not, why not?

Answer 1

The second method is wrong. $T$ is a random variable. While $E\int_0^{T}|B_s|^{2}ds$ is a number, $\int_0^{T}E|B_s|^{2}ds$ is a random variable.