Removing the + c from the antiderivativeWhile I was first learning about integrals and antiderivatives, I took for granted that, for example, ∫ 2 x d x = x 2 + c. But if the constant can be any number, that means that ∫ 2 x d x = x 2 = x 2 + 1 which doesn't make any sense. I understand that the constant is usually just left as c, but it is implied that it can be any number, right? We don't define the square root of 4 to be 2 and -2 and we don't define arcsin(0) to be 0, 2 π, 4 πIf you try to define ∫ d d x ( f ( x ) ) d x = f ( x ), a problem arises. Mainly, if f(x) is something like x 3 + 4. The 4 gets lost when taking the derivative so we can't be sure of what a function is if we only know its derivative.A potential solution to this problem would be to define a function S(f) such that it takes a function as its input and outputs a function stripped of its constant e.g. S ( x 3 + 4 ) = x 3 . If this is done, the antiderivative can be defined as follows: ∫ d d x ( f ( x ) ) d x = S ( f ( x ) ) d d x ( ∫ f ( x ) d x ) = f ( x )Is it possible to define S(f)? If not, is there any other way to define the antiderivative so that it doesn't have a constant?

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Removing the   +  c from the antiderivativeWhile I was first learning about integrals and antiderivatives, I took for granted that, for example,   ∫  2  x  d  x  =      x    2    +  c. But if the constant can be any number, that means that   ∫  2  x  d  x  =      x    2    =      x    2    +  1 which doesn&#039;t make any sense. I understand that the constant is usually just left as c, but it is implied that it can be any number, right? We don&#039;t define the square root of 4 to be 2 and -2 and we don&#039;t define arcsin(0) to be 0,   2  π,   4  πIf you try to define   ∫      d          d      x        (  f  (  x  )  )  d  x  =  f  (  x  ), a problem arises. Mainly, if f(x) is something like       x    3    +  4. The 4 gets lost when taking the derivative so we can&#039;t be sure of what a function is if we only know its derivative.A potential solution to this problem would be to define a function S(f) such that it takes a function as its input and outputs a function stripped of its constant e.g.   S  (      x    3    +  4  )  =      x    3  . If this is done, the antiderivative can be defined as follows:  ∫      d          d      x        (  f  (  x  )  )  d  x  =  S  (  f  (  x  )  )      d          d      x        (  ∫  f  (  x  )  d  x  )  =  f  (  x  )Is it possible to define S(f)? If not, is there any other way to define the antiderivative so that it doesn&#039;t have a constant?

Christina Ward · Accepted Answer

Step 1The integral, while commonly thought of as an inverse to the differentiation operator, is not really a &quot;true&quot; inverse.Just like       f    (    x    )    =          x      2       is, without any extra constraints, technically is not invertible. It lacks injectivity.       f    (    x    )    =    f    (    y    )   does not imply that       x    =    y  . If you want an inverse - you must either resort to multi-valued functions, or you must restrict the domain you are interested in. If we do the latter, then indeed            f              −        1              (    x    )    =          x      In exactly the same way,                     d        f                    d        x              =                  d        g                    d        x            does not imply       f    (    x    )    =    g    (    x    )  . The differential operator clearly lacks injectivity, and you gave concrete examples             d              d        x                    (              x        2            +      1      )        =          d              d        x                    (              x        2            )       but obviously             x      2        +    1    ≠          x      2             ∀         x    ∈          R      .Step 2Making the argument that you can set c to be whatever you want in two different contexts and claiming the results are still equal is an invalid step. It&#039;s like saying             2      2        =    (    −    2          )      2        =    4  , so       2    =    −    2   - injectivity does not hold.Now, when it comes to evaluating definite integrals - it doesn&#039;t matter which anti-derivative you pick. The Fundamental Theorem of Calculus simply states that if f(x) is continuous on [a,b] then            ∫              a                    b              f    (    x    )    d    x    =    F    (    b    )    −    F    (    a    )  where F(x) is any function satisfying             d              d        x                    (      F      (      x      )      )        =    f    (    x    )  . So it doesn&#039;t matter what constant you specify. And it&#039;s easy to see why:            (      F      (      b      )      +      c      )        −          (      F      (      a      )      +      c      )        =    F    (    b    )    −    F    (    a    )  To summarise: you cannot do anything about the missing information (unless in context you can Algebraically find out the value of the       +    c   constant) and it&#039;s not a bug. It&#039;s a feature! (Sorry for the terrible overused Computer Science reference).

Removing the + c from the antiderivative While I was first learning about integrals and an

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