Variance of an Itô integral

Idea. An Itô integral has mean zero, and the Itô isometry turns its second moment into an ordinary integral.

Step 1: mean. $\mathbb{E}[I]=0$, as for any Itô integral of an adapted, square-integrable process.

Step 2: isometry. $$\mathrm{Var}(I)={\int }_0^1\mathbb{E}[W_t^2]dt={\int }_0^{1}tdt=\frac{1}{2}.$$

Check with Itô's formula. $d(W_t^2)=2W_{t}d{W}_t+dt$, so $$I=\frac{W_1^2-1}{2}.$$ Since $W_1^2$ has variance 2 (it is chi-square with one degree of freedom), $\mathrm{Var}(I)=\frac{2}{4}=\frac{1}{2}$.

The trap. Ordinary calculus would give ${\int }_0^{1}WdW=\frac{W_1^2}{2}$, whose mean is $\frac{1}{2}$, not 0. The $-\frac{1}{2}$ is the Itô correction, coming from the quadratic variation of Brownian motion.