# Riemann Sums

• Oct 16th 2007, 05:54 AM
Wyau
Riemann Sums
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.
• Oct 16th 2007, 08:12 AM
galactus
Let's use the right endpoint method:

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

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

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

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

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

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

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

Now, take the limit:

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

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