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Integral problem

  1. Aug 11, 2014 #1
    Is there a non-zero continuous function, f, R -> R such that ∫f(x) dx over (c, c^2+1) = 0, for all c in R? Hope this makes sense. I've been trying to find a more formal way of approaching this rather than just visualizing it with graphs. Any thoughts?

    Many thanks.
     
  2. jcsd
  3. Aug 11, 2014 #2

    maajdl

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    Try to see if you can built a primitive F(c) such that F(c²+1)=F(c) .
    If you can find one, then f is its derivative.

    I guess it might be useful to observe that c²+1 > c .
    Defining F in the interval [0,1] might be all that is necessary.
    Will F be continuous, will its derivative be continuous, ???

    Did you give the full statement of the question?

    Try to graph such a function.
    Give it a try.
     
    Last edited: Aug 11, 2014
  4. Aug 11, 2014 #3

    HallsofIvy

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    Generally, the "zero-function" is f(x)= 0 for all x so that "non- zero" means "not equal to 0 for at least one x". But we can have f(x)= 0 for x= 1, non-zero for x not equal to 0, and its Riemann integral, [itex]\int_a^b f(x)dx[/itex], from any a to any b, is non-zero. If you use the Lebesque function, then we can have f(x) equal to zero on any set of measure 0 and have [itex]\int_a^b f(x)d\mu[/itex] non-zero.
     
  5. Aug 11, 2014 #4

    disregardthat

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    following up on maajdl's comment, the existence of such a function f is equivalent to the existence of a strictly increasing or decreasing differentiable function F : R --> R such that F(c^2+1) = F(c).

    But, it is easily seen that no such function can exist, because the derivative of F(x^2+1) is equal to the derivative of F(x), which means that 2xf(x^2+1) = f(x). Plugging in x = 0 shows that f(0) = 0. Assuming you mean non-zero as in f can not take the value 0.
     
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