Antiderivative of absolute value functionIf I need to find a antiderivative function of a function, for example f ( x ) = | x 2 − 2 x | Can I state that F ( x ) = { x 3 / 3 − x 2 for x &lt; 0 , x &gt; 2 , − x 3 / 3 + x 2 for 0 ≤ x ≤ 2 Or am I not allowed to separate the function into parts like that, and then perform the integral on each part separately? I am concerned with points between the separated parts.If I am not allowed, how can I solve such problems?

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Antiderivative of absolute value functionIf I need to find a antiderivative function of a function, for example  f  (  x  )  =      |          x      2        −    2    x    |  Can I state that  F  (  x  )  =      {                                        x            3                                /                    3          −                      x            2                                     for           x          &amp;lt;          0          ,          x          &amp;gt;          2          ,                                      −                      x            3                                /                    3          +                      x            2                                     for           0          ≤          x          ≤          2                        Or am I not allowed to separate the function into parts like that, and then perform the integral on each part separately? I am concerned with points between the separated parts.If I am not allowed, how can I solve such problems?

fugprurgeil · Accepted Answer

Step 1Since the function is continuous over       R  , you just need to find one antiderivative and the others will differ from it by an additive constant.What antiderivative? The fundamental theorem of calculus provides one! Set   F  (  x  )  =      ∫    0    x        |        t    2    −  2  t      |      d  t and this will be it. Why 0 as the “starting point”? Also 2 would have been fine, other starting points are less well-behaved, because we have   f  (  t  )  =      {                                        t            2                    −          2          t                          t          ≤          0                                      2          t          −                      t            2                                    0          ≤          t          ≤          2                                                  t            2                    −          2          t                          t          ≥          2                        Step 2Hence, for   x  ≤  0, we have  F  (  x  )  =      ∫    0    x    (      t    2    −  2  t  )    d  t  =      1    3        x    3    −      x    2  For   0  ≤  x  ≤  2, we have  F  (  x  )  =      ∫    0    x    (  2  t  −      t    2    )    d  t  =      x    2    −      1    3        x    3  For   x  ≥  2, we have  F  (  x  )  =  F  (  2  )  +      ∫    2    x    (      t    2    −  2  t  )    d  t  =      1    3        x    3    −      x    2    +      8    3

Aganippe76 · Accepted Answer

Explanation:The method is fine. But when you do that, you need to make sure that F is continuous, in particular at the separating points. Right now it’s continuous at 0, but not at 2, so you need to add an appropriate constant in the case   x  &amp;gt;  2

Antiderivative of absolute value function If I need to find a antiderivative function of a function

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