Simple question about antiderivative?So this has been confusing me a lot. Let f ( x ) = x 2 and let F ( x ) = ∫ 1 x f ( x ) d x . Then F ( 1 ) = 0, obviously, but the antiderivative of f (which is the same as ∫ 1 x f ( x ) d x ) is F ( x ) = x 3 / 3, so F ( 1 ) = 1 / 3, not 0?What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as F ( x ) = x 3 / 3 + C, and in this example F ( x ) = x 3 / 3 − 1 / 3, x 3 / 3 is just a special antiderivative where C = 0

Question

Simple question about antiderivative?So this has been confusing me a lot. Let   f  (  x  )  =      x    2   and let   F  (  x  )  =            ∫              1                    x              f    (    x    )              d        x  . Then   F  (  1  )  =  0, obviously, but the antiderivative of f (which is the same as             ∫              1                    x              f    (    x    )              d        x  ) is   F  (  x  )  =      x    3        /    3, so   F  (  1  )  =  1      /    3, not 0?What caused the confusion is that I forgot that a function has infinitely many anti derivatives; they are defined in this case as   F  (  x  )  =      x    3        /    3  +  C, and in this example   F  (  x  )  =      x    3        /    3  −  1      /    3,       x    3        /    3 is just a special antiderivative where   C  =  0

treccinair · Accepted Answer

Explanation:It does not matter what C is equal to.The antiderivative of f is  F  (  x  )  =      ∫    0    x    f  (  t  )  d  t  =      ∫    0    x        t    2    d  t  =  (            t      3        3    +  C  )            |        0    x    =  (            x      3        3    +  C  )  −  (            0      3        3    +  C  )  =            x      3        3    ∴  F  (  1  )  =      ∫    0    1    f  (  t  )  d  t  =  1      /    3

therightwomanwf · Accepted Answer

Step 1The integral       ∫    1    x        x    2      d  x is equal to                                    x          3                3            |        1    x    .This is       1    3        x    3    −      1    3  . Note that       1    3        x    3    −      1    3   is 0 at   x  =  1, as you pointed out it should be.Step 2Remark: When we are finding definite integrals with limits of integration that involve variables, it is best to use a different &quot;dummy&quot; variable of integration. Things may be clearer if you write  F  (  x  )  =      ∫    1    x        t    2      d  t  .While in principle the notation you used is not wrong, it is, for good reason, frowned on.Note also that talking about &quot;the&quot; antiderivative of a function f(t) is not quite right. The function has infinitely many (very closely related) antiderivatives.