I am now studying stochastic process and Ito Integral of Brownian motion. We know $$\int_0^t{B_s}dB_s=\frac{1}{2}B^2_t-\frac{1}{2}t, \text{in } L^2(\Omega)$$
I know the derivation but still really wonder how to "intuitively" explain there is an extra $\frac{1}{2}t$ compared to deterministic integral $\int{x}dx=\frac{1}{2}x^2$.
My first guess is this term is coming from the quadratic variation of Brownian motion. In usual cases (e.g. differentiable functions), the quadratic variation is zero, and hence there is no extra term. Is my guess right?