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Integration subscript question

  1. Apr 27, 2013 #1
    Simple Integration question

    Hey: I am doing a practice exam question and it gives me two families of integrals, the first one being:

    Cn= int(pi,0) xn cos(x) dx int n>= 0
    It has asked for me to do integrate this as C0 I have assumed this means n = 0.
    Do I change the n to a zero before or after integrating?

    I can't find anywhere in my learning centre that tells me this. No urgency, this is just extra study for the exam in a few weeks. I may have additional questions that follow on this, but the integration itself seems straight forward, so do most of the future questions. Thanks in advance.
     
    Last edited: Apr 27, 2013
  2. jcsd
  3. Apr 27, 2013 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hey Intricacy! Welcome to PF! :smile:
    You mean Cn is defined as ∫0π xn cos(x) dx ?

    Then C0 = ∫0π x0 cos(x) dx
     
  4. Apr 27, 2013 #3
    What does x underscript mean?
     
  5. Apr 27, 2013 #4

    Mark44

    Staff: Mentor

    Writing an index as a subscript is a way to indicate multiple variables using only a single letter. For example, x0, x1, and x2 represent three different values.

    Subscripts are often used to represent sequences of numbers or the coefficients of polynomials of arbitrary degree, as in p(x) = c0 + c1x + c2x2 + ... + cnxn.
    Here the coefficients of the terms are the numbers {c0, c1, c2, ... , cn}.

    This is confusing. As tiny-tim already asked, do you mean
    $$C_n = \int_0^{\pi} x_n~cos(x)~dx \text{?}$$

    Or do you mean
    $$C_n = \int_0^{\pi} x^n~cos(x)~dx \text{?}$$


     
  6. Apr 28, 2013 #5
    Oh, I am so sorry for that. Yes I meant x^n, so sorry.

    Thank you for that as well TinyTim. So I can turn the x^0 straight to 1 prior to integration?
    Cn=∫π0 x^n cos(x) dx? ----> C0=∫π0 cos(x)
     
  7. Apr 28, 2013 #6

    tiny-tim

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    Hi Intricacy! :wink:
    Yes … that's the definition of Co. :smile:
     
  8. Apr 28, 2013 #7
    Thank you Tiny-Tim. Sorry for the obviously stupid question :P.
     
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