Explanation:
(B) : \(\int_{-2}^{1}|[x-1]| d x=\)
\(-2\lt x\lt 1\)
\(-2-1\lt \mathrm{x}-1\lt 1-1\)
\(-3\lt \mathrm{x}-1\lt 0\)
\(\therefore \quad[\mathrm{x}-1]=-3,-2,-1\)
\(|[\mathrm{x}-1]|=3,2,1\)
\(\int_{-2}^{1}|[x-1]| d x =\int_{-2}^{+1} 3 d x+\int_{-1}^{0} 2 d x+\int_{0}^{1} 1 d x\)
\(=3[x]_{-2}^{1}+2[x]_{-1}^{0}+[x]_{0}^{1}\)
\(=3(1)+2(1)+1=6\)