You need to do two integrals, one for the interval [-4,-3] using f(x)=-2x-6 and the other for [-3,0] using f(x)=2x+6. You can do this because f(x)=0 when x=-3, so the function is continuous.
∫[-4,-3](-2x-6)dx+∫[-3,0](2x+6)dx=
(-x²-6x)[-4,-3]+(x²+6x)[-3,0]=
(-9+18-(-16+24))+(0-(9-18))=
9-8+9=10, answer C.