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    Hard*

    I can not do.

    Find c in

    c ∫ exp ( -(2x^2+3x+1)) dx =1, from -infinity to +infinity
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    (Original post by AlmostNotable)
    Hard*

    I can not do.

    Find c in

    c ∫ exp ( -(2x^2+3x+1)) dx =1, from -infinity to +infinity
    this undergrad standard
    substitution throws this I think into exp (-x2)
    I can look at this later tonight or tomorrow morning
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    (Original post by TeeEm)
    this undergrad standard
    substitution throws this I think into exp (-x2)
    I can look at this later tonight or tomorrow morning
    Okay I will go back to figuring this can opener out. Real puzzle.
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    (Original post by AlmostNotable)
    Okay I will go back to figuring this can opener out. Real puzzle.
    It is done the way I suggested ...
    Not too bad, assuming you are an undergrad.
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    (Original post by TeeEm)
    It is done the way I suggested ...
    Not too bad, assuming you are an undergrad.
    I am currently doing Alevels.
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    Completing the square we have:

    \displaystyle \int_{-\infty}^{\infty} e^{-(2x^2+3x+1)}\;{dx}= \int_{-\infty}^{\infty} e^{-2(x+\frac{3}{4})^2+\frac{1}{8}} \;{dx} = e^{1/8}\int_{-\infty}^{\infty} e^{-2(x+\frac{3}{4})^2}\;{dx}

    Let t = \sqrt{2}\left(x+\frac{3}{4} \right), then it's \displaystyle \frac{e^{1/8}}{\sqrt{2}}\int_{-\infty}^{\infty}e^{-t^2}\;{dt} =\frac{e^{1/8}\sqrt{\pi}}{\sqrt{2}}
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    (Original post by AlmostNotable)
    I am currently doing Alevels.
    where does the integral come from?
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    (Original post by TeeEm)
    where does the integral come from?
    The cyberspace.
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    (Original post by AlmostNotable)
    The cyberspace.
    I do not follow but this is undergrad stuff.
    here is my version, but this is beyond A level
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