# Definite Integrals Calculus I

1. Dec 5, 2009

### Wm_Davies

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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?

2. Dec 5, 2009

### lanedance

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

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

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

3. Dec 5, 2009

### Wm_Davies

Ahah, integration properties! Thank you.