Approximating discrete sum by integral

[Kashmir]

I can't understand how this approximation works


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


Can you please help me

 

[andrewkirk]

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.

 

[JFerreira]

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

 

[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?

 

[Kashmir]

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

thank you so much

 

[andrewkirk]

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]

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

Thank you :)