Results 1 to 3 of 3

Math Help - some how forgot

  1. #1
    Junior Member
    Joined
    Feb 2009
    Posts
    49

    some how forgot

    Hey guys n gals

    somehow forgot, but when solving a definite integral say functiong y=x^2 from 0 to 10... how come when you do the integral it doesnt include anything below y=0.... i know why but i just neeed the proof

    and when doing area problems under a curve for example sinx {0, 2pi} the value is 0 but the net area is ???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by mpl06c View Post
    how come when you do the integral it doesnt include anything below y=0.
    I have no idea what this means. Please explain more.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,577
    Thanks
    1418
    Quote Originally Posted by mpl06c View Post
    Hey guys n gals

    somehow forgot, but when solving a definite integral say functiong y=x^2 from 0 to 10... how come when you do the integral it doesnt include anything below y=0.... i know why but i just neeed the proof
    Proof of what? That the square of a real number is never negative? Essentially it is that the product of two positive numbers is positive and that the product of two negative numbers is positive: if you are squaring a number, you are multiplying it by itself so you are always multiplying numbers of the same sign.

    and when doing area problems under a curve for example sinx {0, 2pi} the value is 0 but the net area is ???
    The "area under the curve" normally means the area below the graph and above the x-axis. For x between pi and 2pi, there is no such are because y= sin(x) is below y= 0. I suspect you mean the area bounded by y= sin(x) and y= 0. Because "area" is never negative, you are subtracting "higher value" minus "lower value". From 0 to pi, the higher value is sin(x) but from pi to 2pi, the higher value is 0 so you have to do that as two separate integrals: \int_0^\pi (sin(x)- 0)dx+ \int_\pi^{2\pi} (0- sin(x)dx
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: March 10th 2010, 06:24 AM
  2. I forgot how to do this...help.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 16th 2009, 11:29 AM
  3. I forgot how to do this...
    Posted in the Algebra Forum
    Replies: 3
    Last Post: August 30th 2009, 01:57 PM
  4. Forgot
    Posted in the Algebra Forum
    Replies: 5
    Last Post: August 20th 2009, 10:39 PM
  5. Need help! Forgot how to do this..
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 1st 2009, 01:18 PM

Search Tags


/mathhelpforum @mathhelpforum