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Integration of e

  1. Feb 21, 2017 #1
    1. The problem statement, all variables and given/known data


    $$y= \frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2},~y=0,~x=0,~x=1$$

    2. Relevant equations

    $$Volume=2\pi\int_a^b p(x)h(x)dx$$

    3. The attempt at a solution

    I understand how to do the problem, I'm just having trouble integrating.
    ##h(x)=\frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2},~ p(x)=x##
    And you just plug that into the equation.
    When you integrate ##\frac 1 {\sqrt{2\pi}}e^{ \frac{- x^2} 2}x## , you use the chain rule right? But first you move the constant ##\frac 1 {\sqrt{2\pi}}## out of the integral next to the ##2\pi##
    So it would look like this, ##Volume=2\pi~\frac 1 {\sqrt{2\pi}}~\int_0^1~ xe^{ \frac{- x^2} 2}dx##

    I know the anti derivative of ##e^{ \frac{- x^2} 2} is, e^{ \frac{- x^2} 2}## and then you have to do the anti derivative of ## { \frac{- x^2} 2}## as part of the chain rule right? Well, that is where I get stuck.
     
  2. jcsd
  3. Feb 21, 2017 #2
    Hi Jovy:

    I suggest you investigate the substitution u = (x^2)/2, du = x dx

    Hope this helps.

    Regards,
    Buzz
     
  4. Feb 21, 2017 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, the antiderivative of ##e^{-x^2/2}## is a non-elementary function. In other words, it is not possible to write down the antiderivative in a finite number of terms involving the standard functions. Even if you try to write it down in 1 billion pages of horrible algebra, you cannot do it.

    Fortunately, you don't need the antiderivative of ##e^{-x^2/2}##, but of ##x e^{-x^2/2}## instead, and that is a much different story.
     
  5. Feb 21, 2017 #4

    Mark44

    Staff: Mentor

    What exactly is the problem? All you have given us is an equation in x and y, but you haven't said what it is you're supposed to do?

    From your attempt below, it would seem that you're trying to find the volume of a surface of revolution, but there is no clue above that that's what you need to do.
     
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