# Riemann sum

• Nov 21st 2006, 09:53 PM
viet
Riemann sum
Consider the function $\displaystyle f(x) = \frac {x^2}{3} - 9$
In this problem you will calculate $\displaystyle \int_{0}^{2} ( \frac {x^2}{3} - 9) \,dx$ by using the definition

$\displaystyle \int_{a}^{b} f(x) \,dx = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} f(x_i) \Delta x \right]$

The summation inside the brackets is $\displaystyle R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate $\displaystyle R_n$ for $\displaystyle f(x) = \frac {x^2}{3} - 9$
on the interval [0, 2]

$\displaystyle R_n =$

$\displaystyle \lim_{n \to \infty} R_n =$

i have no clue on how to do this one.
• Nov 22nd 2006, 03:23 AM
topsquark
Quote:

Originally Posted by viet
Consider the function $\displaystyle f(x) = \frac {x^2}{3} - 9$
In this problem you will calculate $\displaystyle \int_{0}^{2} ( \frac {x^2}{3} - 9) \,dx$ by using the definition

$\displaystyle \int_{a}^{b} f(x) \,dx = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} f(x_i) \Delta x \right]$

The summation inside the brackets is $\displaystyle R_n$ which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

Calculate $\displaystyle R_n$ for $\displaystyle f(x) = \frac {x^2}{3} - 9$
on the interval [0, 2]

$\displaystyle R_n =$

$\displaystyle \lim_{n \to \infty} R_n =$

i have no clue on how to do this one.

First the interval has n points, so $\displaystyle \Delta x = \frac{2 - 0}{n} = \frac{2}{n}$. Thus each $\displaystyle x_i = 0 + i \cdot \frac{2}{n} = i \cdot \frac{2}{n}$

So:
$\displaystyle R_n = \sum_{i = 1}^n \left ( \frac{x_i^2}{3} - 9 \right ) \frac{2}{n} = \sum_{i = 1}^n \left ( \frac{ \left ( i \cdot \frac{2}{n} \right )^2 }{3} - 9 \right ) \frac{2}{n}$

$\displaystyle R_n = \sum_{i = 1}^n \left ( \frac{8i^2}{3n^3} - \frac{18}{n} \right ) = \sum_{i = 1}^n \frac{8i^2}{3n^3} - \sum_{i = 1}^n \frac{18}{n}$

$\displaystyle R_n = \frac{8}{3n^3} \sum_{i = 1}^n i^2 - \frac{18}{n} \sum_{i = 1}^n 1$

$\displaystyle R_n = \frac{8}{3n^3} \frac{n(n+1)(2n+1)}{6} - \frac{18}{n} n$

$\displaystyle R_n = \frac{4(2n^2 + 3n + 1)}{9n^2} - 18$

Now:
$\displaystyle \lim_{n \to \infty}R_n = \lim_{n \to \infty} \left ( \frac{4(2n^2 + 3n + 1)}{9n^2} - 18 \right )$

= $\displaystyle \lim_{n \to \infty} \frac{4}{9} \left ( 2 + 3 \frac{1}{n} + \frac{1}{n^2} \right ) - 18$

= $\displaystyle \frac{4}{9} \cdot 2 - 18 = \frac{8}{9} - 18 = \frac{-154}{9}$

-Dan
• Nov 22nd 2006, 03:24 AM
galactus
Right end point: $\displaystyle x_{k}=a+k{\Delta}x$.....[1]

Subdivide the interval [0,2] into n equal parts, then each will have length:

$\displaystyle {\Delta}x=\frac{2-0}{n}=\frac{2}{n}$

From [1]:

$\displaystyle x_{k}=\frac{2k}{n}$

The kth rectangle has area:

$\displaystyle f(x_{k}){\Delta}x=\left(\frac{(\frac{2k}{n})^{2}}{ 3}-9\right){\Delta}x$

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

=$\displaystyle \frac{8k^{2}}{3n^{3}}-\frac{18}{n}$

Sum up the rectangles:

$\displaystyle \frac{8}{3n^{3}}\sum_{k=1}^{n}k^{2}-\sum_{k=1}^{n}\frac{18}{n}$

Remember, the sum of $\displaystyle k^{2}=\frac{n(n+1)(2n+1)}{6}$

Sub into our summation:

$\displaystyle \frac{8}{3n^{3}}\cdot\frac{n(n+1)(2n+1)}{6}-18$

=$\displaystyle \frac{-2(77n^{2}-6n-2)}{9n^{2}}$

Now take the limit:

$\displaystyle \lim_{n\to\infty}\frac{-2(77n^{2}-6n-2)}{9n^{2}}$

$\displaystyle \lim_{n\to\infty}\frac{4}{3n}+\lim_{n\to\infty}\fr ac{4}{9n^{2}}-\underbrace{\frac{154}{9}}$

See the limit and, consequently, your integral?.

EDIT: Ol' Topsquark beat me, but after all that I ain't deleting it:)
• Nov 22nd 2006, 03:47 AM
topsquark
Quote:

Originally Posted by galactus
EDIT: Ol' Topsquark beat me, but after all that I ain't deleting it:)

I don't blame you. I'd've done the same thing. :)

-Dan