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Math Help - integral of exp(-x^2)

  1. #1
    Member moonnightingale's Avatar
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    integral of exp(-x^2)

    what is integral of
    Attached Thumbnails Attached Thumbnails integral of exp(-x^2)-pic.gif  
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  2. #2
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    Re: integral of exp(-x^2)

    Quote Originally Posted by moonnightingale View Post
    what is integral of
    e^{-x^2} does not have an elementry antiderivative. The integral can be calulated using Taylor series.

    e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}

    So we can integrate this term by term to get

    \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}

    This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating.
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  3. #3
    Member moonnightingale's Avatar
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    Re: integral of exp(-x^2)

    Quote Originally Posted by TheEmptySet View Post
    e^{-x^2} does not have an elementry antiderivative. The integral can be calulated using Taylor series.

    e^{-x^2}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}

    So we can integrate this term by term to get



    \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}

    This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating.

    Hi if i solve the question by simpsons rule or by matlab i get answer .745
    but i am unable to get this answer by manual method
    plz explain me manual method by puttiing values
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  4. #4
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    Re: integral of exp(-x^2)

    What do you mean by "manual method"?
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  5. #5
    Member moonnightingale's Avatar
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    Re: integral of exp(-x^2)

    i mean how to solve this by hand
    by standard integration method
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  6. #6
    Member moonnightingale's Avatar
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    Re: integral of exp(-x^2)

    Kindly tell me how u will solve this integration question on paper
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    Re: integral of exp(-x^2)

    Quote Originally Posted by moonnightingale View Post
    Kindly tell me how u will solve this integration question on paper
    You have already been told that this is not expressible a closed form in terms of elementary functions. You have been given a series approximation which will give the numerical value to any precision you require.

    \int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}\\ \\ \phantom{SSSSSSS}=\sum_{n=0}^{N-1}\frac{(-1)^n}{(2n+1)n!} +R_N

    Where |R_N|<\frac{1}{(2N+1)N!}


    CB
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  8. #8
    Member moonnightingale's Avatar
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    Re: integral of exp(-x^2)

    can u tell me answer of my question
    with simpson rule it comes to be
    0.7455
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    Re: integral of exp(-x^2)

    You have been told twice now, by TheEmptySet and Captain Black, that [itex]e^{-x^2}[/tex] does not have an "elementary" anti-derivative.

    If you insist upon such an answer then it is \frac{\sqrt{\pi}}{2} erf(1) where "erf" is the Gauss error function. It is defined as \frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx. Does that help?
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    Re: integral of exp(-x^2)

    Quote Originally Posted by moonnightingale View Post
    can u tell me answer of my question
    with simpson rule it comes to be
    0.7455
    If you multiply it by 2/\sqrt\pi, this integral becomes the error function, for which detailed tables of values are available (for example on the Wikipedia page that I just linked to). This gives erf(1) = 0.8427008. So your integral is that number times \sqrt\pi/2, or approximately 0.7468241.
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  11. #11
    MHF Contributor Amer's Avatar
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    Re: integral of exp(-x^2)

    cant we use that method suppose

    I = \int_{0}^{1} e^{-x^2} \cdot dx

    I^2 = \int_{0}^{1} \int_{0}^{1} e^{-x^2}\cdot e^{-y^2} \; dx \;dy

    then transfer to polar ??
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  12. #12
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    Re: integral of exp(-x^2)

    Quote Originally Posted by Amer View Post
    cant we use that method suppose

    I = \int_{0}^{1} e^{-x^2} \cdot dx

    I^2 = \int_{0}^{1} \int_{0}^{1} e^{-x^2}\cdot e^{-y^2} \; dx \;dy

    then transfer to polar ??
    The problem is you need to integrate over the unit square in polar coordinates and that is very difficult. When we go to infinity we can just integrate over a large circle of radius r.
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