How to Prove an Integral with Continuous Functions and Positive Constraints?

[lap]

Homework Statement

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

Homework Equations

Can you help me to prove this integral ?

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

Can you help me to prove this integral ?

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

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.

 

[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 ?

 

[HallsofIvy]

f(a)(b-a) ≤ ∫ f(x) dx from a to b ≤ f(b)(b-a)

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

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

 

[LCKurtz]

Please quote from what you are responding to.

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.

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 ?

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

 

[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 $\int{M\cdot{g(x)}}dx$?

 

[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

 

[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 ?

 

[LCKurtz]

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

 

[dirk_mec1]

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