Find antiderivative of a piecewise functionConsider the function f ( x ) = { x − 1 2 , 0 ≤ x &lt; 1 x − v − 1 2 , v ≤ x &lt; v + 1 ( v ∈ N ) This function is certainly integrable as it only has a countably many discontinuities. For finding it's antiderivative, can I simply consider the piecewise functions, meaning I would obtain for example F ( x ) = { 1 2 x 2 − 1 2 x , 0 ≤ x &lt; 1 1 2 x 2 − v x − 1 2 x , v ≤ x &lt; v + 1 ( v ∈ N ) or is this procedure incorrect?

Question

Find antiderivative of a piecewise functionConsider the function   f  (  x  )  =      {                            x          −                      1            2                    ,                    0          ≤          x          &amp;lt;          1                                      x          −          v          −                      1            2                    ,                    v          ≤          x          &amp;lt;          v          +          1                     (          v          ∈                      N                    )                        This function is certainly integrable as it only has a countably many discontinuities. For finding it&#039;s antiderivative, can I simply consider the piecewise functions, meaning I would obtain for example  F  (  x  )  =      {                                        1            2                                x            2                    −                      1            2                    x          ,                    0          ≤          x          &amp;lt;          1                                                  1            2                                x            2                    −          v          x          −                      1            2                    x          ,                    v          ≤          x          &amp;lt;          v          +          1                     (          v          ∈                      N                    )                         or is this procedure incorrect?

Jayvion Mclaughlin · Accepted Answer

Step 1The function you describe is a saw tooth curve, where for each   v  =  0  ,  1  ,  2  ,  ⋯ the graph is a straight line from coordinate   (  v  ,  −  1      /    2  ) to   (  v  ,  1      /    2  ). Thus if   F  (  x  )  =      ∫    0    x    f  (  t  )  d  t over the range [0,v] for a positive integer v, you will have   F  (  v  )  =  0. In general if   x  =  v  +  ξ where v is an an integer   0  ,  1  ,  2  ,  ⋯ and   0  ⩽  ξ  &amp;lt;  1, you have                    F        (        x        )                            =                  ∫          0          v                f        (        t        )        d        t        +                  ∫          v          x                f        (        t        )        d        t                                          =        0        +                              [                                          1                2                                                    t              2                        −            v            t            −                                          1                2                                      t            ]                    v                      v            +            ξ                                                            =                              1            2                                    ξ          2                −                              1            2                          ξ            Step 2This can be written using the integer part operator, for   x  ⩾  0,                     F        (        x        )        =                              1            2                          (        x        −        ⌊        x        ⌋                  )          2                −                              1            2                          (        x        −        ⌊        x        ⌋        )             which is not the same as the expression you derived.

Find antiderivative of a piecewise function Consider the function f ( x ) = {

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New Questions in Calculus 1