Riemann Sums

**A5HLEY** · April 4th 2008, 05:17 PM

Calculate Rn for f(x)=-[(x^2)/3]+4 on the interval [0,4] and write the answer as a function of n without any summation signs. Also find lim as n-->infinity of Rn.

I never have trouble calculating actual numbers (ie; R4 or L6, etc) but I always have trouble when I'm given a variable. If someone could work this out for me to put in my notes I would really appreciate it!

**TheEmptySet** · April 4th 2008, 06:32 PM

Originally Posted by A5HLEY

Calculate Rn for f(x)=-[(x^2)/3]+4 on the interval [0,4] and write the answer as a function of n without any summation signs. Also find lim as n-->infinity of Rn.

I never have trouble calculating actual numbers (ie; R4 or L6, etc) but I always have trouble when I'm given a variable. If someone could work this out for me to put in my notes I would really appreciate it!

Here is the jist of it...

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

and

$x_i=a+i \cdot \Delta x=0 +i \Delta x=i \Delta x =i\frac{4}{n}$

So we get the sum is

$R_n = \sum_{i=0}^{n}f(x_i) \Delta x = \sum_{i=0}^{n} \left( \frac{((i \frac{4}{n} )^2)}{3}+4 \right) \frac{4}{n}$

$\frac{64}{3n^3} \underbrace{ \sum_{i=0}^{n} i^2}_{=\frac{n(n+1)(2n+1)}{6}}+ \frac{16}{n} \underbrace{ \sum_{i=0}^{n} 1}_{=n}$

$R_n=\frac{64}{3n^3}\frac{n(n+1)(2n+1)}{6}+\frac{16 }{n}n$

You should be able to simplify and take the limit from here.

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