1. Not finding help here? Sign up for a free 30min 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!

Please check my integration

  1. Oct 4, 2004 #1
    From

    [tex]
    \int_{0}^{L} A^2 sin^2 \frac {n \pi x} {L} dx = 1
    [/tex]

    show that

    [tex]
    A = \sqrt {\frac {2} {L}
    [/tex]



    Here's what I did:

    First bring [tex] A^2 [/tex] out so that I have:

    [tex]
    A^2 \int_{0}^{L} sin^2 \frac {n \pi x} {L} dx = 1
    [/tex]

    Then I use u subsitution for the integral:

    [tex]
    \int sin^2 \frac {n \pi x} {L} dx
    [/tex]

    equals:

    [tex]
    \int \frac {L} {n \pi} sin^2 u du
    [/tex]

    Pulling out the constants I get:

    [tex]
    \frac {L} {n \pi} \int sin^2 u du
    [/tex]

    Which, using an integral table, equals:

    [tex]
    (\frac {L} {n \pi}) (\frac {1} {2} u - \frac {1} {4} sin 2u)
    [/tex]

    Plugging u back in and putting in my limits of integration I get my definite integral:

    [tex]
    (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L}
    [/tex]

    Plugging this back into the original equation I get:

    [tex]
    A^2 (\frac {L} {n \pi}) \left[ \frac {n \pi x} {2L} - \frac {sin 2 n \pi x} {4 L} \right]_{0}^{L} = 1
    [/tex]

    Then solving for the limits of integration I get:

    [tex]
    A^2 (\frac {L} {n \pi}) \left[ (\frac {n \pi L} {2L} - \frac {sin 2 n \pi L} {4 L}) - (0 - 0) \right] = 1
    [/tex]

    Then doing some algebra I get:

    [tex]
    A^2 (\frac {n \pi L^2} {2 L n \pi} - \frac {sin 2 n \pi L^2} {4 L n \pi}) = 1
    [/tex]

    ...then some scratch work...

    [tex]
    sin \pi = 0
    [/tex]

    ...so...

    [tex]
    (A^2) (\frac {n \pi L^2} {2 L n \pi}) = 1
    [/tex]

    After more algebra I get:

    [tex]
    (A^2) (\frac {L} {2}) = 1
    [/tex]

    So then:

    [tex]
    A^2 = \frac {2} {L}
    [/tex]

    And finally:

    [tex]
    A = \sqrt {\frac {2} {L}
    [/tex]

    Q.E.D.


    How does it look.
    Sorry if it seems messy; this was my first time using LaTex.
     
  2. jcsd
  3. Oct 4, 2004 #2
    There's a step around plugging n*pi*x/L back in for u where inexplicably the L in n*pi*x/L was pulled out of the sine function. Was this a typo?
     
  4. Oct 4, 2004 #3
    You should really put parentheses around the variables in the sine function, i.e.
    [tex]
    \Big(\frac {L} {n \pi}\Big) \left[ \frac {n \pi x} {2L} - \frac {\sin (2 n \pi x/L)} {4} \right]_{0}^{L}
    [/tex]

    Everything else looks OK.
     
  5. Oct 4, 2004 #4

    Thank you. I will do that for the paper I hand in.


    vsage: I can't tell which step you mean. Could you clarify? B/c it may not have been a typo.


    Anybody else see any mistakes?
     
  6. Oct 4, 2004 #5
    I was referring to what e(ho0n3 pointed out. You seem to account for the typo though later so it looks great to me.
     
  7. Oct 4, 2004 #6
    Excellent. Thanks for looking.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Please check my integration
  1. Check my answer please (Replies: 1)

Loading...