Results 1 to 6 of 6

Math Help - definite integration

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    2

    definite integration

    keep finding that my answers to the area between equation and curve are "correct", only wrong sign. Happens too frequently to be coincidence. Any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Could you provide a specific example? Area must be a positive quantity - always.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    2
    So if my answer is negative I simply reverse the sign?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    Well, it might be more complicated than that. Let me give you an example, and see how you do. Find the area enclosed between the two curves x^{2} and 2-x^{2} from x=-2 to x=2.

    Here's a plot (which, incidentally, I would always start out by doing. It gives a great overview of the kind of thing you need to do.)

    How would you start?

    [EDIT]: Fixed link. Thanks to NOX Andrew.
    Last edited by Ackbeet; March 15th 2011 at 03:44 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Dec 2009
    Posts
    226
    Your link is broken for me. In case it is also broken for hunt798, here is the link to the plot again: http://www.wolframalpha.com/input/?i=Plot[{x^2,2-x^2},{x,-2,2}]
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,394
    Thanks
    1845
    Sounds to me like you are mixing up which curve is above the other.

    For example, if I were asked to determine the area between the curves y= x^2 and y= x+ 2, the first thing I would is draw (or at least visualize) the graphs, noting that they intersect when y= x^2= x+ 2 so that x^2- x- 2= (x+1)(x-2)= 0. That is they intersect at (-1, 1) and (2, 4). I would also recognize that the graph of y= x^2 is always below the graph of y= x+ 2 so the area between them is given by
    \int_{-1}^2 ((x+ 2)- x^2)dx= \frac{1}{2}
    Had I mistakenly written the integral as
    \int_{-1}^2 (x^2- (x+2))dx
    (perhaps just because I am used to writing polynomials with highest power first!)
    I would get -\frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: September 2nd 2010, 01:27 PM
  2. Definite Integration:
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 14th 2009, 06:18 AM
  3. Definite Integration
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 5th 2009, 12:09 AM
  4. definite integration
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 23rd 2008, 11:55 AM
  5. definite integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 3rd 2007, 05:16 PM

Search Tags


/mathhelpforum @mathhelpforum