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Finding the average value given the graph of the derivative

  1. Jan 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Since there is a graph and many things, I will post the link of my question.
    http://ibhl1-bccalculus.wikispaces.com/file/view/Solutions+to+optional+HL1+practice+for+final+(Larson+text).pdf [Broken]
    Please scroll down to the last page (page 16) and my question is 6c.


    2. Relevant equations
    g(x)=3+∫(2,x)f(t)dt


    3. The attempt at a solution
    Here is what I attempt to do: I try to take the integral of g(x) from 2 to 6 and then divide the solution by 4 to find the average value. However, I need to do a double integration on f(t), which I haven't learn yet. I think my approach is too complicated, so is there another way to do the problem? I also think the answer key provided in the link is wrong, can anybody tell me if it is?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jan 25, 2013 #2

    Dick

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    Yes, it's wrong. The smallest value of g in [2,6] is g(4)=3-pi/2. The given answer (5-pi)/4 is smaller than that. The average can't be smaller than the minimum of the function. As you say, they should be integrating g, looks like they are sort of integrating f and doing a bad job of it.
     
    Last edited by a moderator: May 6, 2017
  4. Jan 25, 2013 #3
    Can you tell me how to integrate g, though, I have trouble with that.
     
  5. Jan 25, 2013 #4

    Dick

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    You can't just read it off the graph. You'd need to actually write down a formula for f(x) on each interval and integrate it. The circle part makes it hard. f(x) on [2,4] is given by -sqrt(1-(x-3)^2). You'd need to integrate that to get g(x) which takes a trig substitution that gives an answer involving arcsins and sqrts. The you need to integrate that to get the average. It looks pretty nasty to me. If it were just line segments it would be manageable. I don't think you would be expected to do this.
     
  6. Jan 25, 2013 #5
    I did that before, but I can only integrate the first part because the second part is pretty complex for my level. I am wondering if there is another way to do this problem. I also seek help from another forum before and the helper tells me the following:
    The mean of g(x) between x=2 and x=6 is = 3 + mean of f(x)

    f(x) has negative value between 2 and 4 with area = (1/2).pi x 1^2
    = pi/2

    f(x) is positive between 4 and 6 and area (1/2) x 2 x 2 = 2

    Total area between 2 and 6 = 2 - pi/2 = 0.4292

    Mean of f(x) = 0.4292/4 = 0.1073

    Mean of g(x) = 3 + 0.1073 = 3.1073 <------

    Yet, I still don't think it is right. Can you please tell if what he did is right?
     
  7. Jan 25, 2013 #6

    Dick

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    Try it out on a simpler example. Let's skip the 3 part. Suppose f(x)=x for x in [0,1]. The average of f(x) in [0,1] is 1/2. g(x)=x^2/2. Check that the average of g(x) is 1/6. Not equal to the average of f(x).
     
  8. Jan 25, 2013 #7
    I just have this thought right now. Is it right if I add all the values from g(2) to g(6) and then divide it by 5 to find the average value?
     
  9. Jan 25, 2013 #8

    Dick

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    They goofed up. They know less about the correct way to solve the problem than you do. You were right, you need to integrate twice. I don't see any easy way to solve this problem. You can approximate the answer by finding some points on g(x) and averaging them. The more points you average the better the approximation, but that still not the real answer. I would move on.
     
  10. Jan 26, 2013 #9
    I managed to do the double integral yesterday and got (56-3pi)/24, which is about 1.94, as the final average. Thank you for all your help.
     
  11. Jan 26, 2013 #10

    Dick

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    Hmm. I got 10/3-3pi/8. About 2.16. But I had to correct a few mistakes while I was doing it. Could still be a few left.
     
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