Let $Y(T)=\int_T B(t) dt$ an integral.
We wold like to evaluate $E(Y(T)^2)$
Now, $Y(T)$ may be Riemann integrated because dt have finite absolute variation and $B(t)$ is continuous. Then we can take that integral as the limit N going to infinite in the usual Riemman sum. Then the expected value of the variable $Y(T)^2$ may be written as a double sum of the expected value of $B(t)B(s)$, that is $\min(t,s)$. Performing the sum one can get $T^3/3$.
In contrat, one may take te Ito calculus as follows. One may represent the variable $Y(T)$ by using the integration by parts formula giving
$d(B(T) T)=t dB(t) + B(t) dt$
It seems that
$Y(T)=B(T) T - \int_T t dB(t)$
correct?
Take the square and use the Ito isometry, then
$E(Y(T)^2)=4/3 T^3$
Where is the point here?