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Establishing a definite integral is btw 12&24

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose f(2)=7, f(4)=1, and f'(x)< 0 for all x. Assuming f^(-1) is differentiable everywhere, establish that

    12 < Integral from 1 to 7, of f^(-1)(x)dx < 24


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I do not know where to begin... =/
     
  2. jcsd
  3. Oct 14, 2009 #2

    ideasrule

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    For this problem, you need to think about the graphical interpretation of an integral as the area under a graph. Note also that since f(2)=7, f^-1 (7) =2, and ditto for the other equation.

    Don't get caught up with the 12 and 24. Think about what you would answer if I asked you what values the integral should be between.
     
  4. Oct 14, 2009 #3

    Mark44

    Staff: Mentor

    Start with a graph of y = f-1(x) between 1 and 7. You know that f(2) = 7 and that f(4) = 1. Can you get some points on the graph of f-1? You also know that f'(x) < 0 for all x.
     
  5. Oct 14, 2009 #4
    I made a sort of rectangle when I graphed. I have the points (1,2), (1,4), (7,4) and (7,2).
    From this, i see that the area below 2 is 12, and the area below 4 is 24..
     
  6. Oct 14, 2009 #5

    ideasrule

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    Good! You also know that f'(x) < 0, which means f^-1(x) decreases as x increases, and that means...
     
  7. Oct 14, 2009 #6
    That means.. that f^-1 will eventually be 3. To the left of that point though, it will have to be less than 4, and to the right of that point, it will hve to be greater than 2. ANd... this is why f^-1 cant be equal to the areas of the rectangle that its in..?
     
  8. Oct 16, 2009 #7
    Or is this last part i did wrong?
     
  9. Oct 16, 2009 #8

    ideasrule

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    Your post isn't quite clear, but I think you got it. If you draw out the rectangle bounded by y=2, then you draw out the rectangle bounded by y=4, you'll see that the area under the graph has to be larger than the former and smaller than the latter. There's no way you can twist, turn, bend, or morph the graph so that the area under it is larger than 24 (the larger rectangle's area) or smaller than 12 (the smaller rectangle's area).
     
  10. Oct 16, 2009 #9

    lurflurf

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    use the fact that if
    m<f<M when a<x<b
    then
    [tex]m (b-a)<\int_a^b f dx<M(b-a)[/tex]
    along with some other basic facts
    in particular find m and M so that
    m<f^(-1)(x)<M when 1<x<7
     
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