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Homework Help: 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.
     
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