Show that two primitives/ antiderivatives are related via a constantLet I ⊂ R be an interval. A differentiable function F : I → R is called a primitive for the function f : I → R if F ′ ( x ) = f ( x ) for all x ∈ I.Show: If F 1 and F 2 are two primitives for f on I then there is a constant C ∈ R such that F 2 ≡ F 1 + C, i.e. F 2 ( x ) = F 1 ( x ) + C for all x ∈ I.What we have covered so far is the formal definition of the derivative in terms of the limit: lim h → 0 F 1 ( x + h ) − F 1 ( x ) h = f ( x ) lim h → 0 F 2 ( x + h ) − F 2 ( x ) h = f ( x )I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant 'appears'.We normally just define F 2 = F 1 + C and then it would follow immediately that these two functions have the same derivative.

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Show that two primitives/ antiderivatives are related via a constantLet   I  ⊂  R be an interval. A differentiable function   F  :  I  →  R is called a primitive for the function   f  :  I  →      R   if       F    ′    (  x  )  =  f  (  x  ) for all   x  ∈  I.Show: If       F    1   and       F    2   are two primitives for f on I then there is a constant    C  ∈      R   such that       F    2    ≡      F    1    +  C, i.e.       F    2    (  x  )  =      F    1    (  x  )  +  C for all   x  ∈  I.What we have covered so far is the formal definition of the derivative in terms of the limit:      lim          h      →      0                          F        1            (      x      +      h      )      −              F        1            (      x      )        h    =  f  (  x  )      lim          h      →      0                          F        2            (      x      +      h      )      −              F        2            (      x      )        h    =  f  (  x  )I do not see how I can prove firsthand that these two functions differ by a constant. I know it to be true from my pre-calculus and calculus experience, but how would one make the argument from an analysis point of view, can someone give me a hint as to where this constant &#039;appears&#039;.We normally just define       F    2    =      F    1    +  C and then it would follow immediately that these two functions have the same derivative.

candelo6a · Accepted Answer

Step 1Let   G  (  x  )  =      F    1    −      F    2    ..We observe that       G    ′    (  x  )  =  0. We claim this means that G(x) is constant.Suppose on the contrary the function is non-constant this means there must exist   a  ,  b  ∈  I and where   a  &amp;lt;  b such that   G  (  a  )  ≠  G  (  b  ).Step 2We consider [a,b]. Now We apply the mean value theorem to this interval and the function G(x). We get that there must exist a c such that: Which is a contradiction.      G    ′    (  c  )  =            G      (      b      )      −      G      (      a      )              b      −      a        ∧      G    ′    (  c  )  =  0  ,            G      (      b      )      −      G      (      a      )              b      −      a        =  0    ⟹    G  (  a  )  =  G  (  b  )  .We conclude that G(x) is constant. Thus,       F    2   and       F    1   differ by a constant   ◻.

Alannah Short · Accepted Answer

Explanation:If       F    1    ,      F    2   are primitives of f, by linearity of differentiation,  (      F    1    (  x  )  −      F    2    (  x  )      )    ′    =      F    1    ′    (  x  )  −      F    2    ′    (  x  )  =  f  (  x  )  −  f  (  x  )  =  0.Remains to prove that the antiderivative of 0 is a constant function.

Show that two primitives/ antiderivatives are related via a constant Let I ⊂

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