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Thread: need help integral theorems

  1. #1
    Junior Member
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    need help integral theorems

    State true and false
    1>If f and g are contineous and f(x)>= g(x) for a<x<b, then integral(a to b)f(x)dx>=integral(a tob)g(x)dx
    2>If f and g are contineous and f(x)>= g(x) for a<x<b, then f '(x)>=g'(x) for a<x<b
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  2. #2
    MHF Contributor

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    #1 is a theorem. You should see if you could prove it. It is actually very important.


    #2: On [0,1] consider $\displaystyle f(x) = 2 - x^2 \quad \& \quad g(x) = x^2 . $
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  3. #3
    Super Member

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    Hello, Gracy!

    State true and false.

    (1) If $\displaystyle f$ and $\displaystyle g$ are continuous and $\displaystyle f(x) \geq g(x)$ for $\displaystyle a<x<b$,
    then: .$\displaystyle \int^b_a f(x)\,dx\:\geq \:\int^b_ag(x)\,dx$

    I would say True . . .
    Code:
            |   *
            |   :*        :
            |   :  *      :
            |   :     *   :
            |   :         * f(x)
            |   :         :
            |   :         o g(x)
            |   :      o  :
            |   :  o      :
            |   o         :
          --+---+---------+--
            |   a         b

    If $\displaystyle f(x)$ is always "above" $\displaystyle g(x)$ on $\displaystyle (a,\,b)$
    . . then the area under $\displaystyle f(x)$ will be $\displaystyle \geq$ the area under $\displaystyle g(x).$



    (2) If $\displaystyle f$ and $\displaystyle g$ are continuous and $\displaystyle f(x) \geq g(x)$ for $\displaystyle a<x<b$,
    then: .$\displaystyle f '(x) \geq g'(x)$ for $\displaystyle a<x<b$

    This is False . . .
    Code:
            |   *
            |   :*        :
            |   :  *      :
            |   :     *   :
            |   :         * f(x)
            |   :         :
            |   :         o g(x)
            |   :      o  :
            |   :  o      :
            |   o         :
          --+---+---------+--
            |   a         b

    While $\displaystyle f(x)$ is above $\displaystyle g(x)$, we see that
    . . $\displaystyle f'(x)$ is negative while $\displaystyle g'(x)$ is positive.

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