Proof: Antiderivative of a function differ by a constantLet f(x) a function defined at I ⊆ R and assume that F(x) and G(x) are the antiderivatives of f(x) in I, so there is a c such that for all x ∈ I, F ( x ) = G ( x ) + c.Let us define H ( X ) = F ( x ) − G ( X ) therefore H ′ ( X ) = F ′ ( x ) − G ′ ( X ) = f ( x ) − f ( x ) = 0. Which theorem should be used to finalize the proof?

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Proof: Antiderivative of a function differ by a constantLet f(x) a function defined at   I  ⊆      R   and assume that F(x) and G(x) are the antiderivatives of f(x) in I, so there is a c such that for all   x  ∈  I,   F  (  x  )  =  G  (  x  )  +  c.Let us define   H  (  X  )  =  F  (  x  )  −  G  (  X  ) therefore       H    ′    (  X  )  =      F    ′    (  x  )  −      G    ′    (  X  )  =  f  (  x  )  −  f  (  x  )  =  0. Which theorem should be used to finalize the proof?

Ari Jacobs · Accepted Answer

Step 1This is only true if I is an interval.Let       x    1    ,      x    2    ∈  I, with       x    1    &amp;lt;      x    2  ; then there is   ξ  ∈  (      x    1    ,      x    2    ) with  D  (      x    1    )  −  D  (      x    2    )  =  (      x    2    −      x    1    )      D    ′    (  ξ  )where   D  (  x  )  =  F  (  x  )  −  G  (  x  ). This is the mean value theorem applied to   [      x    1    ,      x    2    ]  ⊆  I.Since       D    ′    (  ξ  )  =      F    ′    (  ξ  )  −      G    ′    (  ξ  )  =  0Step 2More generally, if a differentiable function defined over an interval has zero derivative, then the function is constant.The standard counterexample is   f  (  x  )  =  arctan  ⁡  x  +  arctan  ⁡      1    x   defined on   (  −  ∞  ,  0  )  ∪  (  0  ,  ∞  ). It has everywhere zero derivative, but it is not constant.

hisyhauttaq84w · Accepted Answer

Explanation:First, you need that I is an interval, I assume this is intended. If not it is just not true. (Abstractly, you want the domain to be connected.)Then use the Mean Value Theorem.If H is non-constant there are a,b such that   H  (  a  )  ≠  H  (  b  ), so by MVT you get a contradiction to H′ identically 0.

Proof: Antiderivative of a function differ by a constant Let f(x) a function defined at I &#

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