1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does integrating this work?

  1. Apr 14, 2010 #1
    question:

    Consider the gaussian distribution:

    [tex] p(x) = Ae^(-\lambda (x-a)^2) [/tex]

    (a) use the equation, [tex] 1={\int_{-\infty}^{\infty}} p(x)dx [/tex]

    (b) find <x>, <x^2> and [tex]\sigma[/tex]

    ------------------------------------------------------

    a) if i take (x-a) to be u,

    [tex] 1=\int_{-\infty}^\infty Ae^(-\lambda(x-a)^2)dx[/tex]
    =
    [tex] \int_{-\infty}^\infty Ae^(-\lambda(u)^2)dx [/tex] (not sure if this is right, when i substitute in u, is dx, supposed to be replaced by du?

    why? because of the integration limits? idk understand why please explain,

    - after i get told why that occurs,

    i need help integrating this
    [tex]{\int_{-\infty}^{\infty}} Ae^(-\lambda(u)^2)du [/tex]
     
  2. jcsd
  3. Apr 14, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    Assuming that a is a constant...

    Well, start with your replacement...

    [tex]u=x-a[/tex]

    then differentiate it with respect to x.

    [tex]\frac{du}{dx}=1[/tex]

    And multiply both sides by dx (you can do this).

    [tex]du=dx[/tex]

    So du=dx!
     
  4. Apr 15, 2010 #3
    my first question is how to solve for A and a

    I've been given that i need to figure out A first, before a

    helllp
     
  5. Apr 15, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Are you given a value for lambda? That would really help if you were trying to solve the integral...

    However, first thing I'd do, assuming A is constant, is factor it out of the integral. After all, the resulting integral is solvable, and would be even more solvable if we had a value lambda (and I hope it's a square number too).
     
  6. Apr 15, 2010 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    By the way, with LaTex, put everything you want grouped together, such as an exponent, in { }.

    [ tex ]e^{-\lambda (x- a)^2} [ tex ]
    gives
    [tex]e^{-\lamda (x- a)^2}[/tex]
     
  7. Apr 15, 2010 #6
    Don't worry about integrating that one.

    Instead, try integrating
    [tex]{\int_{-\infty}^{\infty}} (u+a)Ae^{-\lambda(u)^2}du [/tex]
    and
    [tex]{\int_{-\infty}^{\infty}} (u+a)^{2}Ae^{-\lambda(u)^2}du [/tex]
    If you integrate by parts (and you may need to do this more than once, to get rid of the "u+a" coefficients), you'll get (some expression)*(the integral you can't do, which is equal to 1)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How does integrating this work?
Loading...