# 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.

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?!

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?!

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