# Thread: finding area using the limit of a riemann sum

1. ## finding area using the limit of a riemann sum

The sum is ∑ i=1 to n of $(2\frac{2i}{n} +1)*\frac{2}{n}$

multiplying it out:
(*formatting is weird, there should be an n and i=1 around the
∑ and the limit goes all the way down)

lim
n-> infinity ∑ $\frac{8i}{n^2} +\frac{2}{n}$

n-> infinity ∑ $\frac{8}{n^2}$∑i + $\frac{2}{n}$

n-> infinity ∑ $\frac{8}{n^2}[\frac{n(n+1)}{2}] + \frac{2}{n}$

= $\frac{8}{2}$

$=4$

2. Originally Posted by hazecraze

The sum is ∑ i=1 to n of $(2\frac{2i}{n} +1)*\frac{2}{n}$

multiplying it out:
(*formatting is weird, there should be an n and i=1 around the ∑ and the limit goes all the way down)

lim
n-> infinity ∑ $\frac{8i}{n^2} +\frac{2}{n}$

n-> infinity ∑ $\frac{8}{n^2}$∑i + $\frac{2}{n}$

n-> infinity ∑ $\frac{8}{n^2}[\frac{n(n+1)}{2}] + \frac{2}{n}$

= $\frac{8}{2}$

$=4$

What happend to the sum

$\sum_{i=1}^{n}\frac{2}{n}$

You left this out of your calculation!

3. So when it says ∑ 1= n, I plug that in to get $\frac{2}{1}=2$? That + the 4 does make 6.

4. Originally Posted by hazecraze
So when it says ∑ 1= n, I plug that in to get $\frac{2}{1}=2$? That + the 4 does make 6.

That is correct.