Calculus and Integrals

Let: Definite integral f(x) dx=1 from x=3 to 12, Definite integral f(x) dx=3 from x=3 to 6
Definite integral f(x) dx=3 from x=9 to 12

I found the definite integral f(x) dx=-5 from x=6 to 9
Now I need to find the definite integral (1f(x) - 3) dx=? from x=9 to 6
I know it is probably one of those easy ones but I kept going about it and kept coming up with the wrong answer, that is, according to the website that the homework is on. Any help is greatly appreciated.
Thanks

∫[3.12] f(x) dx=1
∫[3.6] f(x) dx=3
∫[9.12] f(x) dx=3


∫[3.12] f(x) dx =∫[3.6] f(x) dx +∫[6.9] f(x) dx +∫[9.12] f(x) dx
1 =3 +∫[6.9] f(x) dx +3
∫[6.9] f(x) dx=1-3-3=-5

∫[9.6] f(x) dx=-∫[6.9] f(x) dx=5

∫[9.6] (f(x) -3) dx=∫[9.6] f(x) dx -3∫[9.6] dx=5 - 3(6-9)=14

∫ f(x) - 3dx = ∫ f(x) dx - ∫ 3 dx