Converting an Integral to a Rieman Sum

  • Thread starter smize
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  • #1
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Main Question or Discussion Point

I know I should know this, but how would one convert a typical integral into a Rieman Sum?

0n sinx + x dx for whatever n.

for example.
 

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  • #2
chiro
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I know I should know this, but how would one convert a typical integral into a Rieman Sum?

0n sinx + x dx for whatever n.

for example.
This function happens to be Riemann Integrable with closed form anti-derivatives: try using the fundamental theorem of calculus to get a closed form expression in terms of G(n) - G(0) where G(x) is the appropriate anti-derivative.
 
  • #3
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I know I should know this, but how would one convert a typical integral into a Rieman Sum?

0n sinx + x dx for whatever n.

for example.

Well, since the function [itex]\,f(x)=\sin x + x\,[/itex] is continuous everywhere, it is Riemann integrable in any finite

interval, and we can choose any partition for it we want, for example the partition
[tex]x_0=0\,,\,x_1=\frac{n}{k}\,,\,x_2=\frac{2n}{k},...,x_k=\frac{kn}{k}=n[/tex]
for the interval [itex]\,[0,n]\,[/itex] , thus

[tex]\int_0^n (\sin x +x)dx=\lim_{k\to\infty}\frac{1}{k}\sum_{i=1}^k \left( \sin \frac{in}{k}+\frac{in}{k} \right)[/tex]

DonAntonio
 
  • #4
Bacle2
Science Advisor
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To add to Don Antonio's comment:

1) Partition your domain of integration [a,b] into a collection a=x0,

x1,....,xn=b .

2)Select a point xi* in each (xi-1,xi).

3)Form the sum Ʃi=1,..,Nf(xi*)(xi-xi-1)

In your case, f(xi*)=xi*+sin(xi*)
 

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