1. We need to evaluate $I=\int_0^2 \left|\,|x-1|-x\right|\,dx.$

2. Break the interval at the point where $|x-1|$ changes form, i.e. at $x=1$.

Case 1: $0\le x\le 1$

Here, $|x-1|=1-x.$ So the integrand becomes $\left|(1-x)-x\right|=|1-2x|.$ This changes sign at $x=\tfrac12$.

So, $\int_0^1 |1-2x|\,dx = \int_0^{1/2}(1-2x)\,dx+\int_{1/2}^1(2x-1)\,dx.$ Now, $\int_0^{1/2}(1-2x)\,dx=\left[x-x^2\right]_0^{1/2}=\frac12-\frac14=\frac14,$ and $\int_{1/2}^1(2x-1)\,dx=\left[x^2-x\right]_{1/2}^1=0-\left(\frac14-\frac12\right)=\frac14.$ Thus, $\int_0^1 |1-2x|\,dx=\frac14+\frac14=\frac12.$

Case 2: $1\le x\le 2$

Here, $|x-1|=x-1.$ So the integrand becomes $|(x-1)-x|=|-1|=1.$ Hence, $\int_1^2 1\,dx=1.$

3. Add both parts: $I=\frac12+1=\frac32.$

4. Therefore, $\boxed{\frac32}$ which as a decimal is $1.50$.