If ∫ 0 on bottom 2 on top, f(x)dx=-3

and ∫ 0 on bottom 5 on top, f(x)dx=7
then ∫ 2 on bottom 5 on top, [4f(x)-1]dx=

A.13 B.15 C.37 D. 39

Please help im confused

∫[2,5] 4f-1 dx

= ∫[2,5] 4f dx - ∫[2,5] 1 dx
= 4∫[2,5] f(x) dx - ∫[2,5] 1 dx
Now, you know that ∫[0,5] = ∫[0,2] + ∫[2,5]
So, ∫[2,5] = ∫[0,5] - ∫[0,2]
Now we can say that
4∫[2,5] f(x) dx - ∫[2,5] 1 dx
= 4(∫[0,5] f(x) dx - ∫[0,2] f(x) dx) - 3
= 4(7 - (-3)) - 3
= 40-3
= 37

Go back and review the linear properties of integrals. They are just sums, so they can be multiplied, added, etc. This problem just works with the fact that the area under a curve can be split into pieces.