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How do I approximate the integral

  1. Apr 24, 2005 #1
    How do I approximate the intergral of [tex]e^{-x^2}[/tex] over some interval?
  2. jcsd
  3. Apr 24, 2005 #2
    [tex]e^{r} = \sum_{n=0}^\infty \frac{r^n}{n!}[/tex]

    Substitute r = -(x^2).

    [tex]e^{-x^2} = \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n}}{n!}[/tex]

    Antidifferentiate series over x.

    [tex]\int {e^{-x^2}}dx = \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)n!} = \sum_{n=0}^\infty \frac{(-1)^{n}x^{2n+1}}{(2n+1)n!}[/tex]

    You can use the last expression with some finite upper value. In other words,

    [tex]\sum_{i=0}^k \frac{(-1)^{i}x^{2i+1}}{(2i+1)i!}[/tex]

    will yield an estimate of the desired antiderivative, with (k+1) being the number of terms involved.

    [Edit: Sorry! Initially, I differentiated the series, instead of antidifferentiating.]
    Last edited by a moderator: Apr 25, 2005
  4. Apr 25, 2005 #3


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    U differentiated the series,u should have integrated it.

    [tex]\int e^{-x^{2}} \ dx=\frac{\sqrt{\pi}}{2}\mbox{erf}\left(x\right) + C [/tex]

    and the error function is tabulated...

  5. Apr 25, 2005 #4
    Yeah, I realized that this morning.

    Sorry, it was kind of late.
  6. Apr 25, 2005 #5
    Thanks to both
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