# How to prove this integral ?

1. May 11, 2014

### lap

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

Let f and g be continuous fuctions on [a,b]. Moreover g(x) > 0 for all x belongs to [a,b].
Show that there is a number c belongs to [a,b] such that

∫ f(x)g(x)dx from a to b = f(c)*∫ g(x)dx from a to b

2. Relevant equations

Can you help me to prove this integral ?

3. The attempt at a solution

I knew that f(c) = 1/( b - a) ∫ f(x)dx from a to b but ∫ f(x)g(x) dx is not eaual to
∫f(x)dx * ∫g(x)dx. I've also tried integration by part but it doesn't work

2. May 12, 2014

### LCKurtz

Hint: Since $f$ is continuous on $[a,b]$, f has a min $m$ and max $M$ on the interval: $m\le f(x)\le M$. Start with that.

3. May 12, 2014

### lap

f(a)(b-a) ≤ ∫ f(x) dx from a to b ≤ f(b)(b-a)
But then how to get the result f(c)∫ g(x)dx from a to b ?

4. May 12, 2014

### HallsofIvy

Staff Emeritus
This is certainly NOT true unless you are assuming that f is an increasing function- which you are not given in this problem.

5. May 12, 2014

### LCKurtz

Please quote from what you are responding to.

Is that supposed to be a response to me? If so, why don't you think about what my hint says about your problem.

6. May 12, 2014

### NihilTico

lap, using LCKurtz' notation, consider any function $f(x)$ on the closed interval $[a,b]$. I think we will both agree that any such function has a min and a max. Not knowing what this function is, we do know that the set of values that $f(x)$ takes on $[a,b]$ has a value at some $A$ in $[a,b]$ and some $B$ in $[a,b]$ such that $f(A)$ is a least upper bound and $f(B)$ is a greatest lower bound. This means that $f(B)$ is less than or equal to all other elements of $f([a,b])$ and $f(A)$ is greater than or equal to all other elements of $f([a,b])$, this is just the extreme value theorem (I'm pretty sure that's what it is called), as you might recall from your first year of calculus. What LCKurtz is asking is that you consider the case where the function $f(x)=m$ and the case where $f(x)=M$, in these cases, we can easily see the inequality that LCKurtz gave $m\le{f(x)}\le{M}$. Because of the properties of this minimum and maximum, what do you think that this says about $\int{m\cdot{g(x)}}dx$ and $\int{M\cdot{g(x)}}dx$?

Last edited: May 12, 2014
7. May 12, 2014

### benorin

$\int{m\cdot{g(x)}}dx\leq \int f(x)g(x) dx\leq\int{M\cdot{g(x)}}dx$ is the next step

8. May 12, 2014

### lap

Thank you very much for all the reply.
So, f(x) became a constant with min value and max value ?
And this is nothing to do with the mean value theorem for integrals ?

9. May 13, 2014

### LCKurtz

The question is whether or not you can figure out what it has to do with your problem.

10. May 13, 2014

### dirk_mec1

Hint: divide all sides by $$\int_a^b g(x)\ dx$$