# Thread:

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

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

EDIT: posted by accident. (Would appreciate it if a mod would delete this)

2. Originally Posted by janvdl
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

$R_{n} = \frac{27}{n^3} \left( \frac{ n(n+1)(2n+1) }{6} \right( + \frac{3}{n} (n)$
what is your question? how thy changed the sums to those formulas? those are identities. see the identities section here

it should not be hard to find the proofs of these identities floating around the internet if you're interested
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