Riemann Sums

**Wyau** · October 16th 2007, 05:54 AM

How do you find the area under the curve using Riemann sums with

f(x) = 9 - x^2, a=0 b=4

How are you supposed to set it up so that the constant doesn't become insignificant when you take the limit? Thank you.

**galactus** · October 16th 2007, 08:12 AM

Let's use the right endpoint method:

${\Delta}x=\frac{b-a}{n}=\frac{4-0}{n}=\frac{4}{n}$

$c_{1}=a+k({\Delta}x)=0+\frac{4k}{n}$

$(9-(\frac{4k}{n})^{2})(\frac{4}{n})=\frac{36}{n}-\frac{64k^{2}}{n^{3}}$

$\frac{36}{n}\sum_{k=1}^{n}(1)-\frac{64}{n^{3}}\sum_{k=1}^{n}k^{2}$

But, $k^{2}=\frac{n(n+1)(2n+1)}{6}$

$36-\left(\frac{64}{n^{3}}(\frac{n(n+1)(2n+1)}{6})\rig ht)$

$=\frac{-32}{n}-\frac{32}{3n^{2}}+\frac{44}{3}$

Now, take the limit:

$-32\lim_{n\rightarrow{\infty}}\frac{1}{n}-\frac{32}{3}\lim_{n\rightarrow{\infty}}\frac{1}{n^ {2}}+\frac{44}{3}$

See what the limit is?. That's your solution.

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