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Homework Help: Average Value of a Function - temperature

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data
    If a cup of tea has temperature 98C in a room where the temperature is 20C, then according to Newton's Law of Cooling the temperature of the tea after t minutes is
    [tex]$\displaystyle \Large T(t)=20 + 78 e^{-t/50}.$[/tex]
    What is the average temperature of the tea during the first 38 minutes?

    2. Relevant equations
    [tex]$ fave ={1}/{b-a} \int _a^bf(x)dx$[/tex]

    3. The attempt at a solution
    [tex]$ fave ={1}/{98-20} \int _{20}^{98} T(t)dt$[/tex] [tex]${fave} =\frac{1}{78}\int _{20}^{98} 20 + 78 e^{-t/50}dt?$[/tex]
  2. jcsd
  3. Feb 9, 2010 #2


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    Well, no. It's integral from t=0 to t=38. Why are you putting the bounds for T in instead of t? It's an average temperature over a time interval.
  4. Feb 9, 2010 #3
    so [tex]
    ${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt?$

    i dont understand your question.. how should i construct the question?
    in my textbook they wrote like 2 sentences about temperature and i dont get it.. they dont even have an example for this type of questions
  5. Feb 9, 2010 #4


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    That's fine. What's your question? You want to average T(t) from t=0 to t=38. That's correct. You just put the T limits in instead of t. That's wrong.
  6. Feb 9, 2010 #5
    [tex] ${fave} =\frac{1}{38}\int _{0}^{38} 20 + 78 e^{-t/50}dt$ [/tex]
    my problem is to find the antiderivative of this i think.
    would it be:
    [tex] ${fave} =\frac{1}{38} (\int _{0}^{38} 20dt + \int _{0}^{38}78 e^{-t/50}dt )$ [/tex]
    [tex] ${fave} =\frac{1}{38} (20t+C|^3^8 + \int _{0}^{38}78 e^{-t/50}dt )$ [/tex]
    [tex] ${fave} =\frac{1}{38} (760 + 78 \int _{0}^{38}e^{-t/50}dt )$ [/tex]

    let u = -t/50
    dt = -50du
    [tex] ${fave} =\int _{0}^{38}e^{u}dt $ [/tex]
    [tex] ${fave} =\int _{0}^{38}e^{u}(-50)du $ [/tex]
    [tex] ${fave} =(-50)* \int _{0}^{38}e^{u}du $ [/tex]
    [tex] ${fave} =(-50)* e^{-t/50} + C |^3^8 $ [/tex]

    -50 * e^(-38/50) = -23.38

    [tex] ${fave} \frac{1}{38} (760 + 78 * (-23.38) )$ [/tex] = -27.997

    im sure its wrong.. but where did i f.. up?
    Last edited: Feb 9, 2010
  7. Feb 9, 2010 #6


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    Yeah, it's definitely wrong. The mean should be somewhere between 98C and 20C. You are just being sloppy. As to where it happened, I would look at where you have a du integral and you still have t limits. If you change the variable from t to u aren't you supposed to change the limits as well?
  8. Feb 9, 2010 #7
    not quite what you did was forget to subtract e^ (-t/50) from e^0.

    ${fave} =\int _{0}^{38}e^{u}(-50)du $


    evaluates to

    e^ (-38/50) - e ^0
    which is something like -.54
    times your constant which was -78*50/38
    gives you a positive number, somewhere around 55.
    add that to

    ${fave} =\int _{0}^{38}du $

    The correct exact answer is 74.63, which makes perfect sense.
    Last edited: Feb 9, 2010
  9. Feb 9, 2010 #8


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    You are forgetting to change the t limits to u limits as well.
  10. Feb 9, 2010 #9
    cant believe how stupid my mistake was :bugeye: this small e^0 changed everything...
    well.. at least i get how to do this kind of question now! its actually not that hard when i think about it now :rolleyes:

    thanks once more guys :wink:
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