help with a difficult integrals?

Question

I need help with this problem. I got the 1st 3 but I'm stuck on the last one. Suppose that f'(x) is continuous on the closed interval [6,9], that f(x) does not equal 0 on this interval, f(6)=3, f(9)=8, and integral 6 to 9 f(x) dx=3

Using this information find the values of the following definite integrals.
1. integral 6 to 9 f'(x)/(f(x)^2) dx

this is one of the other ones
I got.
integral 1/9 to 1/6 f(1/x)/(x^2) dx and that equaled 3.

Answer 1

Given: f(6) = 3, f(9) = 8

Integral (6 to 9, f(x) dx) = 3

1) Integral (6 to 9, [f'(x)] / [f(x)]^2 ) dx

I'm going to write this in a slightly different manner, to make the upcoming substitution obvious.

Integral (6 to 9, (1/[f(x)]^2) [f'(x)] dx )

To solve this, you have to use substitution.
Let u = f(x). Then
du = f'(x) dx.

{Note: Notice how we have f'(x) dx at the tail end of our integral. This is precisely how our "du" turned out to be.}

When we use substitution, our bounds change.
When x = 6, u = f(6) = 3.
When x = 9, u = f(9) = 8

So now our new integral is

Integral (3 to 8, [1/u^2] du)

Integrating this normally, we have

(-1/u) {evaluated from 3 to 8} =
(-1/8) - (-1/3) = -1/8 + 1/3 = -3/24 + 8/24 = 5/24

As a side note, I'm comfounded as to why I didn't need the information that
Integral (6 to 9, f(x) dx) = 3

Nevertheless, that's my answer.

Answer 2

i have to know what f(x) is. YOu are not telling the function, you are telling the general form