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Need some help discarding/tying together ideas

  1. May 15, 2008 #1
    I have been attempting to make the leap from AP Calc to more theory-based calculus by trying to solve problems. Unfortunately, there are no solutions to the book I'm using so I need some guidance. Here's the problem:

    Let f: [0,1] -> (0,1) be continuous. Show that the equation

    [tex] 2x - \int^{x}_{0}\ f(t)dt = 1 [/tex]

    has one and only one solution in the interval [0,1].

    My first instinct was to try the endpoints to support the claim. If x = 0, we find that 0 = 1, which is evidently not a solution. Trying x = 1, we get

    [tex] \int^{1}_{0}\ f(t)dt = 1 [/tex]
    which appears to be valid.

    Noticing the form of the integral, I decided to differentiate both sides with respect to x, getting

    [tex] 2 - f(x) = 0 \Rightarrow f(x) = 2 [/tex]

    But [tex] f(x) = 2 [/tex] implies that the original equation becomes 2x - 2x = 0 = 1, which is clearly not a solution. Since we can choose any x in the interval (0,1) to reach this false conclusion, there is only one solution and it occurs at x = 1.

    I'm pretty confident that the beginning of my proof justifies that x = 0 cannot be a solution and x = 1 is a solution. But I feel that my reasoning as to why there is no solution on (0,1) is faulty. But then again, my reasoning/method of attack could be entirely faulty.
  2. jcsd
  3. May 16, 2008 #2
    There's another problem unfortunately, which is that since f(x) < 1, that you know that
    [tex]\int_0^1 f(t) \, dt < \int_0^1 1 \, dt = 1[/tex]
    Therefore, we cannot have a solution at x=1.

    Similarly, as you've proven, you can't have a solution at x = 0, so none of endpoints can be solutions.

    The manipulations to show that there can be no solution at 0 or 1 are valid. However, your technique of taking the derivative is not valid. Instead of showing that there is no solution in (0,1), you have instead shown that there is not a range of points that are solutions. The reason why this is what shown that is before you can take the derivative, you have to assume that the equation is true for some range of numbers over which you can be sure the equation works (i.e., that it works for all value in some range (a,b)). All you're looking for is a single number though, so this technique will not work.
  4. May 16, 2008 #3
    Ok thanks, I see why the endpoints don't work. But then from the problem statement, there is a solution to the equation in the interval (0,1). Is there a specific theorem I can apply here?

    *EDIT* Btw, this is the fundamental theorem section in Problem Solving through Problems, second problem.
    Last edited: May 16, 2008
  5. May 17, 2008 #4
    Sorry, you'll have to do some calculations here to figure this one out.

    Here, big hint:

    Let [tex]g(x) = 2x - \int_0^x f(x) \, dx[/tex]
    Note that g(0) = 0
    What is g(1)? You can't find an exact answer, but you might be able to put in in some range.

    Using that, you should be able to conclude that there is some x in (0,1) so that g(x) = 1.

    Now why do you know what there can't be two different solutions?
  6. May 18, 2008 #5
    Hmmm, I think when you writing the LHS as a function helped a lot. But following your hints,

    g(0) = 0, g(1) > 1. By the intermediate value theorem, there is a value c such that g(c) = 1. Now to show that this is the only solution on [0,1], we take the derivative

    [tex] g'(x) = 2 - f(x) [/tex]

    Since f(x) has range (0,1), g'(x) has range (1,2). Therefore g'(x) is always positive, so g(x) is strictly increasing on the interval (0,1). But this means that g(x) can only have a value of 1 once on [0,1], since we have shown it cannot be 1 at x = 0 or x = 1.
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