- #1

- 78

- 1

∫

_{0}

^{n}sinx + x dx for whatever n.

for example.

- Thread starter smize
- Start date

- #1

- 78

- 1

∫

for example.

- #2

chiro

Science Advisor

- 4,790

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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.

∫_{0}^{n}sinx + x dx for whatever n.

for example.

- #3

- 606

- 1

∫_{0}^{n}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

- 1,089

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

x

2)Select a point x

3)Form the sum Ʃ

In your case, f(x

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