Finding antiderivative by Maclaurin seriesI want to find ∫ p ( x ) d x where: p ( x ) = e − x 2 2 2 π ..This function can not be manually computed so I am using Maclaurin series to find the antiderivative. I believe the Maclaurin series for this function would be: 1 2 π ∑ n = 0 ∞ − x 2 2 n n ! .The bounds of integration would be from -1 to 1. How would I create a Maclaurin series for the antiderivative from this?

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Finding antiderivative by Maclaurin seriesI want to find   ∫  p  (  x  )  d  x where:  p  (  x  )  =            e                                    −                          x              2                                2                            2      π        ..This function can not be manually computed so I am using Maclaurin series to find the antiderivative. I believe the Maclaurin series for this function would be:      1          2      π            ∑          n      =      0        ∞                                −                      x            2                          2            n              n      !        .The bounds of integration would be from -1 to 1. How would I create a Maclaurin series for the antiderivative from this?

agentbangsterfhes2 · Accepted Answer

Explanation:We can integrate the Maclaurin Series term by term as follows.  2      ∫          0        1        1          2      π            ∑          n      =      0        ∞                                (                                      −                      x            2                          2                                          )                          n                    n      !                  2      π            ∑          n      =      0        ∞        ∫          0        1              (      −      1              )        n                    x                  2          n                                    2        n            n      !                  2      π            ∑          n      =      0        ∞              (      −      1              )        n                    x                  2          n          +          1                                    2        n            n      !      (      2      n      +      1      )                          |                    0        1              2      π            ∑          n      =      0        ∞              (      −      1              )        n                            2        n            n      !      (      2      n      +      1      )        ≈  0.682689492From this series, we get an approximation of the true value.

Finding antiderivative by Maclaurin series I want to find ∫ p ( x ) d

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