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Comparison of Riemann integral to accumulation function

  1. Nov 22, 2011 #1
    Let f:[0,1]→ℝ be an increasing function. Show that for all x in (0,1],

    [tex]\frac{1}{x}\int_{0}^{x}f (t) \,dt \le \int_{0}^{1}f (t) \,dt [/tex]


    So by working backwards I got to trying to show that [itex](1-x)\int_{0}^{1}f (t) \,dt \le \int_{x}^{1}f (t) \,dt [/itex]. While I know both sides are equal at x=1, the LHS is decreasing, and the RHS is continuous, I haven't been able to find a reason to say the inequality is true (probably nothing there).

    Should I pass to considering upper and lower sums, or is there some fancy way to use integration by parts here? I haven't come across anything compelling in all my random scratch work.
     
  2. jcsd
  3. Nov 22, 2011 #2

    lanedance

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    how about something like this...

    f is increasing so
    f(a)<f(b) for all a<b (strictly increasing?)

    now say f(x) = c, can you show
    [tex](1-x)c<\int_x^1 f(t) dt [/tex]
    [tex]\int_0^x f(t) dt < xc [/tex]

    the second gives
    [tex]\frac{1}{x}\int_0^x f(t) dt < c [/tex]
     
    Last edited: Nov 22, 2011
  4. Nov 22, 2011 #3

    lanedance

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    also it may help to think of it like this

    [itex] \frac{1}{x}\int_{0}^{x}f (t) \,dt [/itex] is the average value of the function over the interval [0,x]

    [itex]\int_{0}^{1}f (t) \,dt [/itex] is the average value of the function over the interval [0,1]

    As the function is increasing, so will its average increase over a larger interval
     
  5. Nov 23, 2011 #4
    Thanks for the intuition. Now I was able to easily get the inequality

    [tex]\frac{1}{x}\int^{x}_{0}f\,dt \le \frac{1}{1-x}\int^{1}_{x}f\,dt[/tex]

    but I'm not sure how that can show the LHS ≤ [itex]\int^{1}_{0}f\,dt[/itex]. I'm sure I'm being blind and missing something super obvious. :(
     
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