Integral

$\large \displaystyle \int_0^2 |x-1| .dx = \int_0^1 -(x-1) .dx + \int_1^2 (x-1) .dx\\ \large \displaystyle \implies - \left[ \frac{x^2}{2} - x \right]_0^1 + \left[ \frac{x^2}{2} - x \right]_1^2\\ \large \displaystyle = - \left( \frac{1}{2} - 1 - 0 + 0 \right) + \left( \frac{4}{2} - 2 - \frac{1}{2} + 1 \right)\\ \large \displaystyle = \frac{1}{2} + 0 + \frac{1}{2} = \color{#D61F06}{\boxed{1}}.$

since you know there is no curvature in the function, because it is an absolute value function, you can just plot it and find the area under the function as a triangle

Since A=1 units^2, the value of the integral must be 1.

Let $u=x-1 \implies du=dx$ giving $\int_{-1}^{1}{|u|}du = \left[ \frac{1}{2}u|u| \right ]_{-1}^{1} = \frac{1}{2}+\frac{1}{2}=1$