# Finding exact area with the Definite Integral

• Nov 14th 2007, 11:18 PM
janvdl
Finding exact area with the Definite Integral
Okay i was busy trying to figure these things out and then i got stuck.

Now using the Riemann notation to calculate the area under the graph $\displaystyle x^2 + 1$ we get the following:

$\displaystyle R_{6} = \sum_{i = 1}^{6} [ f(x_{i}) \cdot ( \frac{3-0}{6})]$

(I calculated $\displaystyle R_{6} = 14,375$)

They use rectangles under the graph to approximate the area. (6 rectangles and the x-axis value ranges from 0 to 3.)

Then the book goes further and says:

For any number of rectangles under the graph(Still ranging from 0 to 3 on the x-axis):

$\displaystyle R_{n} = \sum_{i = 1}^{n} [ f(x_{i}) \cdot ( \frac{3-0}{n})]$

At the end we arrive at the following:

$\displaystyle R_{n} = \frac{27}{n^3} \sum_{i = 1}^{n} i^2 + \frac{3}{n} \sum_{i = 1}^{n} 1$

Up to there i understand perfectly.

Then from that we get

$\displaystyle R_{n} = \frac{27}{n^3} \left( \frac{ n(n+1)(2n+1) }{6} \right) + \frac{3}{n} (n)$

Now i have no idea why $\displaystyle \sum_{i = 1}^{n} 1 = n$ Shouldn't it just equal $\displaystyle 1$?

And now we arrive at integration(Still for the graph $\displaystyle x^2 + 1$ ):

$\displaystyle \int_{0}^{3} (x^2 + 1) dx = lim_{n \rightarrow \infty} (12 + \frac{27}{2n} + \frac{9}{2n^2} )$

And the second question: Do we always have to use a Riemann sum to arrive at this part: $\displaystyle 12 + \frac{27}{2n} + \frac{9}{2n^2}$ ? What, no shortcut? :D

And could a mod please delete this(Posted by accident): http://www.mathhelpforum.com/math-he...html#post85119
• Nov 14th 2007, 11:26 PM
Jhevon
Quote:

Originally Posted by janvdl
Now i have no idea why $\displaystyle \sum_{i = 1}^{n} 1 = n$ Shouldn't it just equal $\displaystyle 1$?

that's how summations work. this summation says we are adding 1, n times. by definition, $\displaystyle \sum_{n = 1}^{\infty}c = cn$ for $\displaystyle c$ a constant

you can check the same wikipedia article i suggested in your last post to see why it is defined that way

Quote:

And the second question: Do we always have to use a Riemann sum to arrive at this part: $\displaystyle 12 + \frac{27}{2n} + \frac{9}{2n^2}$ ? What, no shortcut? :D
there are several ways to get to that, or a similar expression. none of these ways are really short though, as long as we are integrating by definition (i'm talking about things like Simpson's rule, the midpoint rule, ...). the shortcut would be the regular integral that you may have seen before, which is

$\displaystyle \int_0^3 \left( x^2 + 1 \right)~dx = \left. \frac 13 x^3 + x \right|_0^3 = 9 + 3 = 12$
• Nov 14th 2007, 11:31 PM
janvdl
Thanks Jhevon, i understand now. (Handshake) :)