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Can you check if it is right?

  1. Sep 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the area of [tex] y= \frac{1}{\cos^2 x} [/tex] from [tex] x=0 [/tex] to [tex] x = \frac{\pi}{2} [/tex].

    2. The attempt at a solution


    [tex] \int \frac{1}{\cos^2 x}[/tex] from 0 to [tex] \frac{\pi}{2} [/tex].
    equals [tex] \tan x [/tex]
    and take the limit from the negative side of [tex] \frac{\pi}{2} [/tex].
    since limit of tan at [tex] \frac{\pi}{2} [/tex] is [tex] \infty [/tex]
    the funtion is divergent.
    so the area dosen't exist.
     
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2
    I was wondering why no one answer this forum, you only have to check if I am right( yes or no).
     
  4. Sep 23, 2007 #3
    The integral is divergent.
     
  5. Sep 23, 2007 #4

    learningphysics

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    Yes, you are right.
     
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