Can we allow f to be undefined at finitely many points in (a,b) when formulating ∫ a b f ( x ) d x = F ( b ) − F ( a ), (F is the antiderivative of f)Let f be a real-valued function on a closed interval [a,b] undefined only at finite points in (a,b). Let F be antiderivative of f. Then: ∫ a b f ( x ) d x = F ( b ) − F ( a )Is the theorem true? How shall we prove it?I ask this because I have read that function undefined at finite points in the interval (a,b) doesn't effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

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Can we allow f to be undefined at finitely many points in (a,b) when formulating       ∫    a    b    f  (  x  )     d  x  =  F  (  b  )  −  F  (  a  ), (F is the antiderivative of f)Let f be a real-valued function on a closed interval [a,b] undefined only at finite points in (a,b). Let F be antiderivative of f. Then:      ∫    a    b    f  (  x  )     d  x  =  F  (  b  )  −  F  (  a  )Is the theorem true? How shall we prove it?I ask this because I have read that function undefined at finite points in the interval (a,b) doesn&#039;t effect integration i.e. we can take the antiderivatives and apply the limits as usual to get the definite integral.

Charlee Gentry · Accepted Answer

Explanation:That statement doesn&#039;t make sense, since f may well not be integrable on [a,b] then. And if it is, the statment is not necessarily true. Take, for instance, the null function f on   [  0  ,  2  ]  ∖  {  1  }. Then                     F        :                    [        0        ,        2        ]        ∖        {        1        }                    ⟶                              R                                          x                    ↦                              {                                                    0                                             if                 x                ∈                [                0                ,                1                )                                                                    1                                             if                 x                ∈                (                1                ,                2                ]                                                                  is an antiderivative of f, but       ∫    0    2    f  (  x  )        d    x  =  0  ≠  1  =  F  (  2  )  −  F  (  0  )  .

Can we allow f to be undefined at finitely many points in (a,b) when formulating <msubsup>

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