finding area using the limit of a riemann sum

**hazecraze** · December 6th 2009, 04:05 PM

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

= $\frac{8}{2}$

$=4$

Answer should be 6.

**TheEmptySet** · December 6th 2009, 04:20 PM

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

= $\frac{8}{2}$

$=4$

Answer should be 6.

What happend to the sum

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

You left this out of your calculation!

**hazecraze** · December 6th 2009, 05:08 PM

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

**TheEmptySet** · December 6th 2009, 05:27 PM

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.

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