Proof that two antiderivatives differ by a constant, quantifier confusionThe proof that two antiderivatives of f differ by a constant is given in several other questions.By the mean value theorem, for all a &lt; b ∈ I, there exists an x ∈ [ a , b ] such that ( F − G ) ′ ( x ) = ( F − G ) ( b ) − ( F − G ) ( a ) b − a = F ′ ( x ) − G ′ ( x ) = f ( x ) − f ( x ) = 0 ⟹ 0 = ( F − G ) ( b ) − ( F − G ) ( a ) b − a ⟹ ( F − G ) ( b ) = ( F − G ) ( a ) ⟹ F ( b ) − G ( b ) = F ( a ) − G ( a )so we can define C := F ( b ) − G ( b ) = F ( a ) − G ( a )But since that's true for all a and b, what happens if we let a = inf I? Then F ( a ) = G ( a ) = 0 ,, which means that F ( b ) = G ( b ) for all b &gt; inf I. But that doesn't seem like it's true.So I suspect I'm confused about quantifiers somewhere. Where am I going wrong?

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Proof that two antiderivatives differ by a constant, quantifier confusionThe proof that two antiderivatives of f differ by a constant is given in several other questions.By the mean value theorem, for all   a  &amp;lt;  b  ∈  I, there exists an   x  ∈  [  a  ,  b  ] such that  (  F  −  G      )    ′    (  x  )  =            (      F      −      G      )      (      b      )      −      (      F      −      G      )      (      a      )              b      −      a          =      F    ′    (  x  )  −      G    ′    (  x  )    =  f  (  x  )  −  f  (  x  )    =  0      ⟹    0  =            (      F      −      G      )      (      b      )      −      (      F      −      G      )      (      a      )              b      −      a            ⟹    (  F  −  G  )  (  b  )  =  (  F  −  G  )  (  a  )      ⟹    F  (  b  )  −  G  (  b  )  =  F  (  a  )  −  G  (  a  )so we can define  C  :=  F  (  b  )  −  G  (  b  )  =  F  (  a  )  −  G  (  a  )But since that&#039;s true for all a and b, what happens if we let   a  =  inf  I? Then   F  (  a  )  =  G  (  a  )  =  0  ,, which means that   F  (  b  )  =  G  (  b  ) for all   b  &amp;gt;  inf  I. But that doesn&#039;t seem like it&#039;s true.So I suspect I&#039;m confused about quantifiers somewhere. Where am I going wrong?

Sasha Pacheco · Accepted Answer

Step 1If f is a continuous function on the interval I and   p  ∈  I, then   F  (  x  )  =      ∫    p    x    f  (  t  )    d  t is an antiderivative of f. Similarly, if   q  ∈  I, then   G  (  x  )  =      ∫    q    x    f  (  t  )    d  tThere is no reason why at any point   a  ∈  I it holds that   F  (  a  )  =  G  (  a  )  =  0.Step 2Also, antiderivatives need not be of the above form. Just to make a simple example, if f is the constant zero function, then       ∫    p    x    f  (  t  )    d  t is the constant zero function for every p, but any constant is an antiderivative.You are misguided by the context. More generally, without reference to antiderivatives,if H is a differentiable function on the interval I and H′ is the constant zero function, then H is constant.Proof. Let   a  ,  b  ∈  I, with   a  &amp;lt;  b. By the mean value theorem there exists   c  ∈  (  a  ,  b  ) such that   H  (  b  )  −  H  (  a  )  =  (  b  −  a  )      H    ′    (  c  )  =  0. Thus   H  (  a  )  =  H  (  b  ). Since a and b are arbitrary, we have that   H  (  x  )  =  H  (  a  ), for every   x  ∈  I

Proof that two antiderivatives differ by a constant, quantifier confusion The proof that two antide

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