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Need help with a simple integral proof

  1. Jul 23, 2013 #1
    f is a positive decreasing function of x, f(n) = an for all n:

    [itex]\int\stackrel{n+1}{1} f(x) dx[/itex] [itex]\leq[/itex] a[itex]_{1}[/itex] + a[itex]_{2}[/itex] + [itex]\cdots[/itex] + a[itex]_{n}[/itex]

    Why is this true? I think I'm missing a fundamental understanding of summation, since the reasoning for this step was left out in my book.

    [edited out some nonsense]

    Thanks for any help!
     
    Last edited: Jul 23, 2013
  2. jcsd
  3. Jul 23, 2013 #2

    hilbert2

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    Write the left side of the inequality as [itex]\int^{2}_{1}f(x)dx+\int^{3}_{2}f(x)dx+\dots\int^{n+1}_{n}f(x)dx[/itex] and the right side as [itex]\int^{2}_{1}a_{1}dx+\int^{3}_{2}a_{2}dx+\dots\int^{n+1}_{n}a_{n}dx[/itex] and show that each term on the LHS is smaller than the corresponding term on the RHS. Note that [itex]\int^{b}_{a}f(x)dx\leq\int^{b}_{a}g(x)dx[/itex] if [itex]f(x)\leq g(x)[/itex] on interval [a,b].
     
  4. Jul 23, 2013 #3
    Alright. I've got this, I think.

    In order to make sense of the series, I should represent it as a piecewise graph with height an and a width Δx, where Δx is always 1. The area of each piece of such a graph will be an*Δx, which is just an.

    If Ʃ an is continuous, etc, then I can compare it to a function, f. f(x) is always going to dip underneath (in the case of decreasing functions) or be equal to the horizontal line y = an for each interval. Thus the area of f(x) is always less than the area of the piecewise graph. And, when calculating the area of f(x), I need to go out one more step, n+1, since the Δx from the piecewise graph steps out one further.

    Thank you very much!
     
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