# Math Help - Finding exact area with the Definite Integral

1. ## 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 $x^2 + 1$ we get the following:

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

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

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):

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

At the end we arrive at the following:

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

Up to there i understand perfectly.

Then from that we get

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

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

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

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

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

And could a mod please delete this(Posted by accident): http://www.mathhelpforum.com/math-he...html#post85119

2. Originally Posted by janvdl
Now i have no idea why $\sum_{i = 1}^{n} 1 = n$ Shouldn't it just equal $1$?
that's how summations work. this summation says we are adding 1, n times. by definition, $\sum_{n = 1}^{\infty}c = cn$ for $c$ a constant

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

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

$\int_0^3 \left( x^2 + 1 \right)~dx = \left. \frac 13 x^3 + x \right|_0^3 = 9 + 3 = 12$

3. Thanks Jhevon, i understand now.