Results 1 to 3 of 3

Math Help - Integration question

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    9

    Integration question

    I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks
    Attached Thumbnails Attached Thumbnails Integration question-equation.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,089
    Thanks
    314
    Sorry, but as written the integral evaluates to infinity, not 1: the value of e^{-a(t-2)} for t \to -\infty is \infty.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by bidii View Post
    I've obtained the anti-derivative, but I'm having problems solving for the constant a from this step. Just to let you know, the expected value of t is 6, so I already have the anti-derivative of that equation but I'm having problems going forward. Also, the lower limit can be considered as starting from 0. Thanks

    For a\neq 0 , \int\limits_{-\infty}^\infty ae^{-a(t-2)}\,dt=-[e^{-a(t-2)}]^\infty_{-\infty}=-\lim_{b\to\infty}e^{-a(b-2)}+\lim_{h\to -\infty}e^{-a(h-2)} . Let's now check each limit separatedly:

    \lim_{b\to\infty}e^{-a(b-2)}=\lim_{b\to\infty}\frac{1}{e^{a(b-2)}} =\left\{\begin{array}{lc}0&,\,if\,\,a>0\\\infty&,\  ,if\,\,a<0\end{array}\right. , and likewise \lim_{h\to -\infty}e^{-a(h-2)}=\lim_{h\to -\infty}\frac{1}{e^{a(h-2)}} =\left\{\begin{array}{lc}\infty&,\,if\,\,a>0\\0&,\  ,if\,\,a<0\end{array}\right.

    According to the above the value of the improper integral can't be 1 at all, since the integral diverges for any value of a.

    I don't understand what you mean by "the expected value of t is 6"...

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Integration question
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: February 24th 2011, 09:44 AM
  2. Replies: 6
    Last Post: July 21st 2010, 06:20 PM
  3. Integration question - 2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 10th 2009, 10:42 PM
  4. integration question help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 17th 2008, 06:48 PM
  5. a question about Integration
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 26th 2006, 01:12 PM

Search Tags


/mathhelpforum @mathhelpforum