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If f:[0,1] -> R is a continuous function, describe f.

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    f:[0,1] [tex]\rightarrow[/tex] R is a continous function such that
    [tex]\int[/tex]f(t)dt (from 0 to x) = [tex]\int[/tex] f(t)dt( from x to 1) for all x[tex]\in[/tex][0,1] .
    Describe f.




    2. Relevant equations
    integral represents area



    3. The attempt at a solution

    what ever the function is, I know that the area under the graph from 0 to x is equal to the area under the graph from x to 1.
    But how else can I describe f?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 7, 2009 #2

    Mark44

    Staff: Mentor

    One possibility is that f(t) = 0 for all t in [0, 1]. If there are other functions, none come to mind.
     
  4. Feb 7, 2009 #3
    Note that the integral is a signed area, so areas under the x-axis are negatively signed. If you draw a picture of a function that is entirely on one side of the axis (positive or negative) and start integrating from 0 to small x, it implies the remaining curve must take a nose-dive in order to account for the small amount of area between 0 and x. On the other hand, if we place x close to 1, it implies the small region between x and 1 must be higher than the previous region in order to account for the larger area, which doesn't agree with the previous scenario (or the function is the 0 function).
    This implies that the function must cross the x-axis between 0 and 1 (or be the 0 function). Can you continue from there?
    Another less geometric and more algebraic approach is to use the fundamental theorem of calculus to note that there exists some function F on [0,1] such that the first integral is F(x) - F(0) and the second integral is F(1) - F(x), which leads to the same conclusion.
     
  5. Feb 7, 2009 #4
    I think so,Thank you
     
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