$\int\limits_{-1}^{1} (|x - 2| + |x|) dx = $

5

$\int\limits_{-1}^{1} (|x - 2| dx + \int\limits_{-1}^{1} |x| dx$

$\int\limits_{-1}^{1} -x + 2~dx + 2\int\limits_{0}^{1} x dx$

even function

|x| = x for x > 0

|x - 2| = $\left\{\begin{array}{l}x-2 \quad x \geq 2 \\ -x+2 \quad x<2\end{array}\right.$

So interval (-1, 1)

⇒ |x - 2| = -x + 2

$=\left[\frac{-x^2}{2}+2 x\right]_{-1}^1+\left[x^2\right]_0^1$

= 4 + 1

= 5 sq. units