Converting an Integral to a Rieman Sum

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.

 

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

 

To add to Don Antonio's comment:

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

x₁,....,x_n=b .

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

3)Form the sum Ʃ_i=1,..,Nf(x_i*)(x_i-x_i-1)

In your case, f(x_i*)=x_i*+sin(x_i*)