Converting an Integral to a Rieman Sum

[smize]

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.

 

[chiro]

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.

 

[DonAntonio]

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 $\,f(x)=\sin x + x\,$ 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
$$x_0=0\,,\,x_1=\frac{n}{k}\,,\,x_2=\frac{2n}{k},...,x_k=\frac{kn}{k}=n$$
for the interval $\,[0,n]\,$ , thus

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

DonAntonio

 

[Bacle2]

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*)