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Simple exercise about integrals

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Let ## f: [0, a]## ---> ## \mathbb{R}## be positive and increasing.
    Prove that the function G, such that:
    ##G(x):= \frac{1}{x} \int_0^x f(t)dt## ## x\in (0,a)##
    is increasing.

    3. The attempt at a solution
    I know that if the first derivative of a function is positive, that function is increasing.
    By definition of primitive function, we have that the derivative of ##\int_0^x f(t)dt## is ##f##, which is positive by hypothesis.
    However, i can't handle that ##\frac{1}{x}##. If I derivate it I get ##\frac{1}{x^2}##, making things worse.
    What should I do with it?
     
  2. jcsd
  3. Dec 26, 2012 #2
    Have you tried to write out a formula for [itex]G'(x)[/itex]? (Apply the rule for the derivative of a product.)
     
  4. Dec 26, 2012 #3
    yes, and i get ##G ' (x)= -\frac{1}{x^2} \int_0^x f(t)dt + \frac{1}{x} f(x)+c ##
    ##= \frac{1}{x}(-\frac{1}{x}\int_0^x f(t)dt+f(x)+c)##
    but there appears a minus, and i don't know how to go on with the proof
     
  5. Dec 26, 2012 #4
    I've had an idea, maybe it's correct:
    Let us consider ##G ' (x)##
    ##\frac{1}{x} ## is a positive number (because x>0 in the hypothesis), so i can do as if it weren't there.
    Now,
    ##-\frac{1}{x} \int_0^x f(t)dt+f(x)+c > 0## if ##f(x)+c > \frac{1}{x} \int_0^x f(t)dt##
    But ##f(x)## is the upper bound, so for the integral mean theorem we have:
    ##f(x) (\frac{1}{x-0}) > \int_0^x f(t)dt> m (\frac{1}{x-0})##
    and therefore:
    ##f(x)+c> (\frac{1}{x})\int_0^x f(t)dt##
    QED.
    or have I made any mistakes?
     
  6. Dec 26, 2012 #5
    For simplicity, let the antiderivative of f be F.
    Then,
    [tex]-\frac{1}{x}\int_0^x f(t)dt+f(x) = -\frac{1}{x} (F(x) - F(0)) + f(x)[/tex]
    We can apply the mean value theorem in reverse:
    There exists a point in the interval [0,x] (lets call it c) such that [tex]F'(c) = f(c) = \frac{F(x) - F(0)}{x-0}[/tex]
    Then,
    [tex]-\frac{1}{x} (F(x) - F(0)) + f(x) = -f(c) + f(x)[/tex]
    But f is an increasing function, so f(x) > f(c) because c < x.
     
  7. Dec 26, 2012 #6
    Great!
    Do you think my version above could work as well?
    Anyway, loads of thanks!
     
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