# Thread: Mean-value theorem for integrals??!

1. ## Mean-value theorem for integrals??!

Mean-value theorem for integrals:

========================================
Suppose that f is continuous on [a,b],Then there is a number c on [a,b] such that

$\int_a^b f(x)dx=(b-a)f(c)$

========================================

My question:

I would like to prove that:

if f is a decreasing function on [a,b],then C less than or equal to (a+b)/2

Can you help me to do that??!

Thanks for any responding

2. Originally Posted by miss_lolitta
I would like to prove that:

if f is a decreasing function on [a,b],then C less than or equal to (a+b)/2

Can you help me to do that??!

Thanks for any responding
Note: You can make the theorem stronger. The function does not need to be continous, it can be integrable.

Now, decreasing function are integrable so the theorem applies.

You can answer your question now.

Hint: Assume not, i.e. C>(a+b)/2 then C*(b-a) will be larger than the value of the integral, because the function is decreasing.

3. Originally Posted by miss_lolitta
Mean-value theorem for integrals:

========================================
Suppose that f is continuous on [a,b],Then there is a number c on [a,b] such that

$\int_a^b f(x)dx=(b-a)f(c)$

========================================

My question:

I would like to prove that:

if f is a decreasing function on [a,b],then C less than or equal to (a+b)/2

Can you help me to do that??!

Thanks for any responding
It looks to me as though its not true.

Consider a=0, b=1, and y1(x)=-x^2+1, and y2(x)=x^2-2x+1.
Both y1 and y2 are decreasing on [0,1].

Then:

integral_{x=0 to 1} y1(x) dx = 2/3

and

integral_{x=0 to 1} y2(x) dx = 1/3.

Then solving the equations:

(b-a)y1(c) =2/3

for c in (0,1) gives c=1/sqrt(3)~=0.58

and again:

(b-a)y2(c) =1/3

for c in (0,1) gives c=1-1/sqrt(3)~=0.42.

In one case we have c>(a+b)/2 and in the other c<(a+b)/2.

RonL

4. After I posted my approach, I was too begining to think it was wrong.

For example, f(x)=0 on [0,1] it is decreasing (if you accept that definition). Certainly there are infinitely many differenent c which make this true.

5. Thanks for all

6. Consider a=0, b=1, and y1(x)=-x^2+1, and y2(x)=x^2-2x+1.
Both y1 and y2 are decreasing on [0,1].

Then:

integral_{x=0 to 1} y1(x) dx = 2/3

and

integral_{x=0 to 1} y2(x) dx = 1/3.

Then solving the equations:

(b-a)y1(c) =2/3

for c in (0,1) gives c=1/sqrt(3)~=0.58

and again:

(b-a)y2(c) =1/3

for c in (0,1) gives c=1-1/sqrt(3)~=0.42.

In one case we have c>(a+b)/2 and in the other c<(a+b)/2.
now,,we deduce that :

c exists at neighbourhood of the center of interval..

now,,if this logic is true ,How can I prove that??!

can someone help me?!

Thanks in advance

7. Originally Posted by miss_lolitta
now,,we deduce that :

c exists at neighbourhood of the center of interval..

now,,if this logic is true ,How can I prove that??!

can someone help me?!

Thanks in advance
Not sure what this means. The two examples I worked can be adapted to move the larger c as close to 1 as I want and the lesser as close to 0 as
I want (at least I think I can).

RonL

8. I think that:

c in the mean -value theorem satisfies :

[tex]c\in((a+b)/2-\epsilon,(a+b)/2+\epsilon)[\math]"just suggesting I'm not sure"..

I would like to figure out if this thinking is true or not??

9. Originally Posted by miss_lolitta
I think that:

c in the mean -value theorem satisfies :

[tex]c\in((a+b)/2-\epsilon,(a+b)/2+\epsilon)[\math]"just suggesting I'm not sure"..

I would like to figure out if this thinking is true or not??
If epsilon = (b-a)/2 yes, but that is the same as saying c in [a,b].

RonL