Results 1 to 3 of 3

Math Help - Finding exact area with the Definite Integral

  1. #1
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by janvdl View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Thanks Jhevon, i understand now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] definite integral that gives area of a region
    Posted in the Calculus Forum
    Replies: 12
    Last Post: October 20th 2011, 09:58 PM
  2. Replies: 7
    Last Post: April 25th 2011, 08:08 AM
  3. Replies: 4
    Last Post: April 13th 2011, 02:08 AM
  4. Area from a definite integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 25th 2008, 01:00 PM
  5. Finding exact area with the Definite Integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 14th 2007, 11:14 PM

Search Tags


/mathhelpforum @mathhelpforum