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Find the average mean of a function

  1. Aug 14, 2009 #1
    1. The problem statement, all variables and given/known data
    The mean daily temperature in degrees Fahrenheit of a certain city on the xth day of the year is approximated by the function
    f(x)= 64 + 27 cos[(2π/365) (x+145)] (note: π = pie)
    Part d) Find the average mean daily temperature

    2. Relevant equations



    3. The attempt at a solution
    I have done part a to c of this problem. But when I got to part d, I really did not know how. I tried to plug it into the calculator and tried to find how to find the mean, but I could not find anything. Please help me to solve this.
     
  2. jcsd
  3. Aug 14, 2009 #2

    MathematicalPhysicist

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    You should calculate sqrt(<f,f>) (<,> is the inner product).

    For trigonometric functions the domain of integration is usually of length 2pi, i.e from 0 to 2pi or from -pi to pi, depending on the problem.

    Here to get appropiate domain of integration use the substitution y=(2pi/365)(x+145) where y ranges from 0 to 2pi, from here you can get the domain of integration for x.

    Does it help?
     
  4. Aug 14, 2009 #3

    CompuChip

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    The average of a function f over an interval [a, b] is given by

    [tex]\operatorname{avg}_{[a,b]} f = \frac{1}{b-a} \int_a^b f(x) \, \mathrm dx[/tex]
     
  5. Aug 14, 2009 #4

    MathematicalPhysicist

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    Yes, obviously I forgot.

    To be more formal, there's a weighted function if I'm not mistake, if you take a weighted average.
     
  6. Aug 16, 2009 #5
    Thank you for helping, but I do not get what you are talking about. This is my summer work for calculus and I'm not in calculus yet, so can you please make it simple.
    What are a, b, d, and x in this problem and what is the symbol that looks like the letter "f" means?
    Thank you so much. I really appreciate your help.
     
    Last edited: Aug 16, 2009
  7. Aug 16, 2009 #6

    CompuChip

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    Ah sorry, you asked for the average in the Calculus & Beyond forum, so I thought you'd seen integration already.

    You can do it without, in that case, by making a graph of the function. Looking at it, what do you notice? Any intuition as to what the average may be?
     
  8. Aug 16, 2009 #7
    It's alright and I am sorry if I posted in the wrong section. Anyway, how do I find the average mean from looking at the graph? I thing (probably wrong) that after pluging the equation into TI83plus, I will push "2nd" and "trace" and "7". In this problem, is the lower limit 0 and upper 365? Or is the lower limit 1 because there is no day zero?

    This is what I got so far. I did the steps that I listed above (using "0" for the lower limit and 365 for the upper limit) and got 23360 then I mutiplied the number by (1/(b-a)) or (1/(365-0)). The answer that I got is 64. Is this correct? Or should I use "1" instead of "0" in the place of "a"?

    Please help me. Thank you.
     
    Last edited: Aug 16, 2009
  9. Aug 17, 2009 #8

    jgens

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    64 looks correct. You can do this problem without integration if you examine the term [itex]27cos((2\pi/365)(x + 145))[/itex]
     
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