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Math Help - Justifying the integration of both sides

  1. #1
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    Justifying the integration of both sides

    How can you prove using the definition of anti-differentiation, that if two functions are equal, then their anti-derivatives must differ by a constant?

    Also can this be used to prove the constant rule for integration??
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  2. #2
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    First, if H'(x) = 0, then by definition of the derivative, it must be that H(x) = constant.

    If you have F and G s.t. F'=G' = f, then (F - G)' = F' - G' = f - f = 0.
    Since (F - G)'(x) = 0, then it must be that (F - G)(x) = constant, or F(x) - G(x) = constant, or F(x) = G(x) + constant.

    That is if both F and G belong to the set of anti-derivatives of f, they may differ by a constant.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Skyrim View Post
    How can you prove using the definition of anti-differentiation, that if two functions are equal, then their anti-derivatives must differ by a constant?

    Also can this be used to prove the constant rule for integration??
    It's by a simple fact that if f:\mathbb{R}\to\mathbb{R} (or any connected subspace of \mathbb{R} in fact) is such that f'\equiv 0 then f is constant. To see this fact let x,y\in\mathbb{R} be arbitrary. By the MVT there exists some \xi\in(x,y) such that f(x)-f(y)=(x-y)f'(\xi)=0 and thus f(x)=f(y). Since x,y were arbitrary the ocnclusion follows.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Skyrim View Post
    How can you prove using the definition of anti-differentiation, that if two functions are equal, then their anti-derivatives must differ by a constant?
    The statement is not true. Choose for example:

    0,1)\cup (2,3)\rightarrow{\mathbb{R}}" alt="F,G0,1)\cup (2,3)\rightarrow{\mathbb{R}}" />

    F(x)=\begin{Bmatrix} 0& \mbox { if }& x\in (0,1)\\1 & \mbox{if}& x\in (2,3)\end{matrix}\quad G(x)=\begin{Bmatrix} 1 & \mbox{ if }& x\in (0,1)\\0 & \mbox{if}& x\in (2,3)\end{matrix}

    Then, F'=G' , however F and G don't differ by a constant.

    The statement is true if the domain of definition for F and G is an interval of \mathbb{R} .

    Hint : Use the Mean Value Theorem.


    Fernando Revilla

    P.S Edited: Sorry, I didnīt see the previous posts. Anyway, perhaps the counter-example could be useful.
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    Quote Originally Posted by FernandoRevilla View Post
    The statement is not true. Choose for example:

    0,1)\cup (2,3)\rightarrow{\mathbb{R}}" alt="F,G0,1)\cup (2,3)\rightarrow{\mathbb{R}}" />

    F(x)=\begin{Bmatrix} 0& \mbox { if }& x\in (0,1)\\1 & \mbox{if}& x\in (2,3)\end{matrix}\quad G(x)=\begin{Bmatrix} 1 & \mbox{ if }& x\in (0,1)\\0 & \mbox{if}& x\in (2,3)\end{matrix}

    Then, F'=G' , however F and G don't differ by a constant.

    The statement is true if the domain of definition for F and G is an interval of \mathbb{R} .

    Hint : Use the Mean Value Theorem.


    Fernando Revilla

    P.S Edited: Sorry, I didnīt see the previous posts. Anyway, perhaps the counter-example could be useful.
    Could you please prove it for me, assuming that indeed F and G lie on the real intervals.
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Skyrim View Post
    Could you please prove it for me, assuming that indeed F and G lie on the real intervals.
    I suppose you meant the first part. The second one has already been proved in this thread.

    For every x\inS=(0,1)\cup (2,3) we verify F'(x)=G'(x)=0 . That is, F and G are primitives of f=0 on S. However:

    (F-G)(x)=\begin{Bmatrix} -1& \mbox { if }& x\in (0,1)\\\;\;1 & \mbox{if}& x\in (2,3)\end{matrix}

    i.e. F-G is a non constant function.


    Fernando Revilla
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