- #1

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##\sum_{k=0}^m\left(\frac{k}{m}\right)^n\approx\int_0^m\left(\frac{x}{m}\right)^ndx\tag{1}##

Can you please help me

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- #1

- 441

- 73

##\sum_{k=0}^m\left(\frac{k}{m}\right)^n\approx\int_0^m\left(\frac{x}{m}\right)^ndx\tag{1}##

Can you please help me

- #2

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Then set n=1, and draw the graph of ##f(x) = (\frac xm)^n## between 0 and ##m##. On the same set of axes draw the series of m rectangles such that the k-th rectangle (k = 0, ..., m-1) is the set of points (x,y) with ##\frac km\le x\le \frac{k+1}m## and ##(\frac xm)^n\le y\le (\frac{x+1}m)^n##.

Then do the same thing on a new set of axes, for ##n=2##.

That should give you a strong sense for why the approximation works.

- #3

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Geogebra can help you see the graphs: https://www.geogebra.org/graphing/sfgvm5zu

- #4

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We first think about the sum as the area under the rectangles having unit width.

Then set n=1, and draw the graph of ##f(x) = (\frac xm)^n## between 0 and ##m##. On the same set of axes draw the series of m rectangles such that the k-th rectangle (k = 0, ..., m-1) is the set of points (x,y) with ##\frac km\le x\le \frac{k+1}m## and ##(\frac xm)^n\le y\le (\frac{x+1}m)^n##.

Then do the same thing on a new set of axes, for ##n=2##.

That should give you a strong sense for why the approximation works.

This area is roughly the integral of the corresponding continuous function.

Am i right?

- #5

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thank you so muchGeogebra can help you see the graphs: https://www.geogebra.org/graphing/sfgvm5zu

- #6

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Yes, that's correct. It is called a Riemann Sum. You can read more about them here, including some nice pictures.We first think about the sum as the area under the rectangles having unit width.

This area is roughly the integral of the corresponding continuous function.

Am i right?

- #7

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Thank you :)Yes, that's correct. It is called a Riemann Sum. You can read more about them here, including some nice pictures.

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