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?

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