Results 1 to 2 of 2

Math Help - Integration

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5

    Integration

    \displaystyle\frac{4L^2}{\pi^2(1+2m)^2}\int_0^L\si  n^2\left(\frac{\pi x(1+2m)}{L}\right)dx

    \displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2}  \int_0^L\left(1-\cos\left(\frac{2\pi x(1+2m)}{L}\right)\right)dx

    \displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2}  \left[x-\frac{L}{2\pi(1+2m)}\sin\left(\frac{2\pi x(1+2m)}{L}\right)\right]_0^L=\frac{2L^3}{\pi^2(1+2m)^2}

    Correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by dwsmith View Post
    \displaystyle\frac{4L^2}{\pi^2(1+2m)^2}\int_0^L\si  n^2\left(\frac{\pi x(1+2m)}{L}\right)dx

    \displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2}  \int_0^L\left(1-\cos\left(\frac{2\pi x(1+2m)}{L}\right)\right)dx

    \displaystyle\Rightarrow\frac{2L^2}{\pi^2(1+2m)^2}  \left[x-\frac{L}{2\pi(1+2m)}\sin\left(\frac{2\pi x(1+2m)}{L}\right)\right]_0^L=\frac{2L^3}{\pi^2(1+2m)^2}

    Correct?
    Yes, since the integral is L/2 (assuming m and integer)

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 3rd 2010, 12:54 AM
  2. Replies: 2
    Last Post: November 2nd 2010, 04:57 AM
  3. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  4. Replies: 2
    Last Post: February 19th 2010, 10:55 AM
  5. Replies: 6
    Last Post: May 25th 2009, 06:58 AM

/mathhelpforum @mathhelpforum