What is the correct solution to \int_{0}^{-2} (9 f(x)- 2)dx?

[Wm_Davies]

Homework Statement

Let
$$\int_{-4}^{2} f(x) dx =9, \int_{-4}^{-2} f(x) dx=2, \int_{0}^{2} f(x)dx =3$$

Find (I have solved this one)
$$\int_{-2}^{0} f(x)dx= 4$$

Find
$$\int_{0}^{-2} (9 f(x)- 2)dx= ?$$

Homework Equations

$$\int_{a}^{b} f(x) dx = F(b) - F(a)$$

The Attempt at a Solution

I don't see where I am doing something wrong.

$$\int_{0}^{-2} (9 f(x)- 2)dx$$

$$-\int_{-2}^{0} (9 f(x)- 2)dx$$

Using the fundamental theorem of calculus.
b=0
a=-2

-[(9b(4) - 2b) - (9a(4) - 2a)]
-[(0(4) - 0) - (9(-2)(4) - 2(-2))]

Solving for that I get

-[-(-72+4)]
-[-(-68)]

Therfore I figure that

$$\int_{0}^{-2} (9 f(x)- 2)dx= -68$$

However the electronic homework says it is incorrect, but why?

 

[lanedance]

may be easier to see if you write it out as
$$\int_{0}^{-2} (9 f(x)- 2)dx$$

$$=-\int_{-2}^{0} (9 f(x)- 2)dx$$

$$=-9\int_{-2}^{0} f(x)dx+\int_{-2}^{0} 2dx$$

 

[Wm_Davies]

may be easier to see if you write it out as

Ahah, integration properties! Thank you.