Integral Application

**Del** · November 11th 2007, 09:25 AM

Consider the function

.

In this problem you will calculate

by using the definition

The summation inside the brackets is

which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate

for

on the interval

and write your answer as a function of

without any summation signs.

I need to find

and

Please help!

**galactus** · November 11th 2007, 04:08 PM

right endpoint method: $x_{i}=a+k{\Delta}x$

${\Delta}x=\frac{3-0}{n}=\frac{3}{n}$

$x_{i}=\frac{3k}{n}$

$f(x_{i}){\Delta}x=\left[\frac{(\frac{3k}{n})^{2}}{4}-9\right]\cdot\frac{3}{n}$

$=\frac{27k^{2}}{4n^{3}}-\frac{27}{n}$

$\sum_{i=1}^{n}f(x_{i}){\Delta}x=\frac{27}{4n^{3}}\ sum_{i=1}^{n}k^{2}-\frac{27}{n}\sum_{i=1}^{n}1$

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

So, we have:

$\frac{27}{8n}+\frac{9}{8n^{2}}-\frac{99}{4}$

Now, take the limit:

$\lim_{n\rightarrow{\infty}}\left[\frac{27}{8n}+\frac{9}{8n^{2}}-\frac{99}{4}\right]$

See, the limit?. You should get the same as if you integrate the 'easy' way.

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