Answer:
Option B
Explanation:
$|x|+|x-1| = \begin{cases}-x-(x-1)=-2x+1, if & x \leq 0\\ x-(x-1) = 1, if & 0\leq x \leq 1\\ x+(x-1) = 2x-1,if&x\geq1\end{cases}$
.'.$\int_{-1}^{3}(|x|+|x-1|)dx $ =
$\int_{-1}^{0}(-2x+1)dx +\int_{0}^{1}1dx+\int_{1}^{3}(2x-1)dx$
= $\left[-x^{2}+x\right]_{-1}^{0}+\left[x\right]_0^1+\left[x^{2}-x\right]_1^3$ =9
