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Math Help - Mean-value theorem for integrals??!

  1. #1
    Junior Member miss_lolitta's Avatar
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    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
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  2. #2
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    Quote Originally Posted by miss_lolitta View Post
    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.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta View Post
    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
    Last edited by CaptainBlack; May 7th 2007 at 01:40 PM.
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    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.
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    Junior Member miss_lolitta's Avatar
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    Thanks for all
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    Junior Member miss_lolitta's Avatar
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    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
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  7. #7
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta View Post
    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
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  8. #8
    Junior Member miss_lolitta's Avatar
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    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??
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by miss_lolitta View Post
    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
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