Evaluating a definite integral of an absolute value function

Since $x-1 \le 0$ on $[0,1]$ and $x - 1 \ge 0$ on $[1,2]$, we split the integral as $\int_0^2 |x-1|\,dx = \int_0^1 (1-x)\,dx + \int_1^2 (x-1)\,dx$. An antiderivative of $1-x$ is $x - \frac{x^2}{2}$, and an antiderivative of $x-1$ is $\frac{x^2}{2} - x$.

By FTC Part 2, $\int_0^1 (1-x)\,dx = \left[x - \frac{x^2}{2}\right]_0^1 = \left(1-\frac12\right)-(0-0) = \frac12$. Similarly, $\int_1^2 (x-1)\,dx = \left[\frac{x^2}{2}-x\right]_1^2 = (2-2)-\left(\frac12-1\right) = \frac12$. Adding these gives $\frac12+\frac12 = 1$.