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 Post subject: FTC - spaces between curvesPosted: Fri, 13 Apr 2012 14:06:25 UTC
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Joined: Fri, 11 Nov 2011 14:43:25 UTC
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Prove that

provided that (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

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 Post subject: Re: FTC - spaces between curvesPosted: Fri, 13 Apr 2012 14:43:12 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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nikniklau wrote:
Prove that

provided that (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

No. It is about estimating integral from above and below (i.e. upper and lower Riemann sum, for suitable partition P).

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 Post subject: Re: FTC - spaces between curvesPosted: Fri, 13 Apr 2012 18:44:28 UTC
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Joined: Fri, 11 Nov 2011 14:43:25 UTC
Posts: 18
outermeasure wrote:
nikniklau wrote:
Prove that

provided that (this is the space between curve 1/x, x axis, x=n and x=n+1 as I calc it)

This is about the FTC. Any hints?

No. It is about estimating integral from above and below (i.e. upper and lower Riemann sum, for suitable partition P).

so the lower sum is and the upper sum is and I have to figure out what the partition is.

That is to find for and for when x in [a,b].
I can find that trying but what is a suitable method for it? It is clear what max and min is for this function. Since x>0 and a<b then f(a)=max and f(b)=min.

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 Post subject: Re: FTC - spaces between curvesPosted: Sat, 14 Apr 2012 01:58:53 UTC
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Quote:
(...) provided that

given in the question. It is there for a reason!

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 Post subject: Re: FTC - spaces between curvesPosted: Sat, 14 Apr 2012 11:28:34 UTC
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Joined: Fri, 11 Nov 2011 14:43:25 UTC
Posts: 18
. That is the space between line , x axis, lines x=1 and x=k (dk=1). How is this connected to the other parts of the inequality? ln(n+1)>1 when n>e...

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 Post subject: Re: FTC - spaces between curvesPosted: Sat, 14 Apr 2012 15:19:51 UTC
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nikniklau wrote:
. That is the space between line , x axis, lines x=1 and x=k (dk=1). How is this connected to the other parts of the inequality? ln(n+1)>1 when n>e...

I think you want [tex]\sum_{k=1}^n\frac{1}{k{/tex] instead.

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 Post subject: Re: FTC - spaces between curvesPosted: Sun, 15 Apr 2012 20:30:11 UTC
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Joined: Fri, 11 Nov 2011 14:43:25 UTC
Posts: 18
Oops...Since dk=1 then this can be
This is ln(n+1) for a = 1 and the inequality stands (as an equality on the left). And then what? Proof by induction?

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 Post subject: Re: FTC - spaces between curvesPosted: Tue, 17 Apr 2012 15:35:27 UTC
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Joined: Fri, 11 Nov 2011 14:43:25 UTC
Posts: 18

Some videos over utube and my knowledge over Riemann sums is kind of better now.
The above sum is the left Riemann Sum. Its first term is 1 and that means f(a)dx = 1 and since dx=1 then f(a)=1 and a=1.
, that means that the function is monotonically decreasing and the left sum is greater that the right sum. I think the inequality is in a good way now, since we can calculate the difference between the left and the right sum.

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