Antiderivative of a functionProve or disprove that f ( x ) = { sin ⁡ ( 1 x ) for x ≠ 0 1 2 for x = 0 has an antiderivative in R .This function satisfies the Darboux's property, but how to say definitely that admits an antiderivative?

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Antiderivative of a functionProve or disprove that  f  (  x  )  =      {                            sin          ⁡          (                      1            x                    )                                      for                           x              ≠              0                                                                        1            2                                                for                           x              =              0                                              has an antiderivative in       R  .This function satisfies the Darboux&#039;s property, but how to say definitely that admits an antiderivative?

Nia Molina · Accepted Answer

Step 1It doesn&#039;t have an antiderivative. Let   g  (  x  )  =      {                            sin          ⁡          (                      1            x                    )                                       if                           x              ≠              0                                                            0                                       if                           x              =              0                        .                                  Then g has an antiderivative. So, if f had one,   f  −  g would have an antiderivative too. But f−g doesn&#039;t satisfy the Darboux property.In order to see why g has an antiderivative, consider the function  h  (  x  )  =      {                                        x            2                    cos          ⁡                      (                          1              x                        )                                     if           x          ≠          0                                      0                           if           x          =          0.                        Step 2Then       h    ′    (  x  )  =      {                            2          x          cos          ⁡                      (                          1              x                        )                    +          sin          ⁡                      (                          1              x                        )                                     if           x          ≠          0                                      0                           if           x          =          0.                        But the function   x  ↦      {                            2          x          cos          ⁡                      (                          1              x                        )                                     if           x          ≠          0                                      0                           if           x          =          0                        is continuous and therefore has an antiderivative. Therefore, g has an antiderivative too.

Antiderivative of a function Prove or disprove that f ( x ) = { <mta

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