Confusion on ambiguity of antiderivatives.Would this statement, ∀ k ∈ R , be true or false? And are these interchangeable under all circumstances? ∫ f ( x ) d x + k = ∫ f ( x ) d xSince we could assign F(x) to either the LHS or RHS, and differentiating F(x) would both result with f(x). However if we assign I = ∫ f ( x ) d x, then I = I + k ⇒ k = 0 even though we already stated "for all k". I think my misunderstanding stems from notation and what they represent. Would appreciate any clarification on this as I could not find a direct take on this flexibility of antiderivatives, especially in such context of equality.

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Confusion on ambiguity of antiderivatives.Would this statement,   ∀  k  ∈      R  , be true or false? And are these interchangeable under all circumstances?  ∫  f  (  x  )  d  x  +  k  =  ∫  f  (  x  )  d  xSince we could assign F(x) to either the LHS or RHS, and differentiating F(x) would both result with f(x). However if we assign   I  =  ∫  f  (  x  )  d  x, then   I  =  I  +  k  ⇒  k  =  0 even though we already stated &quot;for all k&quot;. I think my misunderstanding stems from notation and what they represent. Would appreciate any clarification on this as I could not find a direct take on this flexibility of antiderivatives, especially in such context of equality.

humanistex3 · Accepted Answer

Step 1No, your statement doesn&#039;t hold in general, I think you&#039;re misunderstanding the notation a bit. For a differentiable function F such that       F    ′    =  f we have that   ∫  f  (  x  )  d  x  =  F  (  x  )  +  k.Step 2Notice that functions that differ in a constant have the same derivative, that&#039;s why we consider k when writing the antiderivative. For example take   f  (  x  )  =      x    2   and   g  (  x  )  =      x    2    +  1, then       f    ′    (  x  )  =  2  x  =      g    ′    (  x  ).

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