$\int_0^t W_s \, ds $ has finite variation

Question

Let $W=\{W_t:t \geq 0\}$ be a standard Brownian motion and consider $$X_t=\int_0^t W_s \, ds$$

We define the variation of a process $X_t$ has $$Var[X]_t=\lim_{n\to\infty} \sum_{i=1}^n |X_{t_i}-X_{t_{i-1}}|$$ with $t_i=\frac{t}{n}i$.

How do I compute $Var[X]_t$ ?

In particular, why is it finite?

Answer 1

$$\sum_i^n |X_{t_{i+1}}-X_{t_i}|=\sum_i^n\left|\int_{t_i}^{t_{i+1}} W_sds\right|\leq \sum_i^n\int_{t_i}^{t_{i+1}} |W_s|ds=\int_0^t|W_s|ds<\infty$$