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Can some one give me a proof for this

  1. Feb 16, 2012 #1
    http://imageshack.us/photo/my-images/17/4andfinal.jpg/

    I ve discussed with a lot of people at my university including a teacher but they equate the above equation inorder to get the area under a graph which is sinusoidal for 0 to pi and a zero funtion for pi to 2pi....

    So plz give me a proof so that I can show them.....i.e. their method is mathematically wrong...
     
  2. jcsd
  3. Feb 16, 2012 #2

    micromass

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    Errr, why would they say that??? Did they give any explanation for it?? Maybe you misunderstood what they said??
     
  4. Feb 16, 2012 #3
    The graph was actually of a sinusoidal wave from 0 to pi...and from pi to 2pi it remained zero....making the periodic function for the ac wave...
    Now the problem was that they said to find the equivalent dc current...i.e. to find the average value u use this formula to find the area...

    integrate Asinxdx from o to 2pi...
    whereas I insisted it will be
    integrate Asinxdx from 0 to pi...as the sinusoidal graph is only for this interval

    and then divide the area by 2pi....

    and then they said u can split the limits...i.e.
    integrate sinxdx from o to 2pi = integrate sinxdx from o to pi + integrate 0dx from pi to 2pi...

    this is where I objected that is equality sign is not correct ....it doesn't satisfy mathematical laws....
    am I correct..??
    and plz if I am....help me ....prove it to them...
     
  5. Feb 16, 2012 #4

    micromass

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    You're right. Writing

    [tex]\int_0^{2\pi}\sin(x)dx=\int_0^\pi\sin(x)dx+ \int_\pi^{2\pi}0dx[/tex]

    is obviously not correct.

    However, if they wrote

    [tex]f(x)=\left\{\begin{array}{l} \sin(x)~\text{if}~0\leq x\leq \pi\\ 0~\text{if}~\pi\leq x\leq 2\pi \end{array}\right.[/tex]

    then writing

    [tex]\int_0^{2\pi}f(x)dx=\int_0^\pi\sin(x)dx+\int_\pi^{2\pi}0dx[/tex]

    is correct.
     
  6. Feb 16, 2012 #5
    thanks alott.....micromass......hope this satisfies them...
     
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