Correct option is (4) 14

\( \int_{-1}^1\left(\frac{x^2+3}{2}+(x-1)\right) d x+\int_1^2((3-x)-(x-1)) d x \)
\(\Rightarrow \frac{x^3}{6}+\frac{3 x}{2}+\frac{x^2}{2}-\left.x\right|_{-1} ^1+3 x-\frac{x^2}{2}-\frac{x^2}{2}+\left.x\right|_1 ^2 \)
\(A=\frac{7}{3} \Rightarrow 6 A=14\)