# Approximating discrete sum by integral

• I
• Kashmir

#### Kashmir

I can't understand how this approximation works

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

Choose a moderate value of m, not too small, say m=20.
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.

WWGD and Kashmir
WWGD and Kashmir
Choose a moderate value of m, not too small, say m=20.
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.
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?

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

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