Results 1 to 6 of 6

Math Help - Trapezoidal Rule

  1. #1
    Junior Member Misa-Campo's Avatar
    Joined
    Jul 2008
    From
    Australia, Sydney
    Posts
    34

    Trapezoidal Rule



    Can i please solution to this question , thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    The four subdivisions are 1 -> 1.5, 1.5 -> 2, 2 -> 2.5 and 2.5 -> 3.

    So you calculate the value of the function at each of the 5 points at the boundaries of these intervals and apply the rule for finding the area of a trapezoid:

    A = x (y_1 + y_2) / 2

    for each trapezoid, then add them all up.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member Misa-Campo's Avatar
    Joined
    Jul 2008
    From
    Australia, Sydney
    Posts
    34
    Sorry, i still dont get it wats the forumula you are using
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Take the first trapezoid. That is, the one that goes from x=1 to x=1.5.

    Thus the x distance is 0.5.

    The y distance at x=1 (which we can call y_1) is f(1) = 16-1=15 (from f(x) = 16-x^2).

    The y distance at x=1.5 (which we can call y_2) is f(1.5) = 16-(1.5)^2 = 16-2.25 = 13.75.


    So the area of the trapezoid = x (y_1 + y_2) / 2 (that is, multiply x by the mean of y_1 and y_2) is 0.5 (16 + 13.75) / 2 = whatever your calculator gives you.

    Do the same thing with all the other intervals between 1.5 and 2.

    Then take all these 4 areas and add them up.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by Misa-Campo View Post


    Can i please solution to this question , thanks
    The integration is from x=1 to x=3. Four sub-intervals are needed, so
    (3 -1)/ 4 = 0.5 per interval.

    We need to find the y-values when x=1, x=1.5, x=2, x=2.5, and x=3
    y = f(x) = 16 -x^2
    So,
    y(1) = 16 -1^2 = 15
    y(1.5) = 16 -(1.5)^2 = 13.75
    y(2) = 16 -2^2 = 12
    y(2.5) = 16 -(2.5)^2 = 9.75
    y(3) = 16 -3^3 = 7


    So by trapezoidal rule,
    INT(1 to 3)[16 -x^2]dx
    = (d/2)[h1 +2h2 +2h3 +2h4 +h5]
    = (0.5 /2)[15 +2(13.75) +2(12) +2(9.75) +7]
    = 23.25 sq.units.

    -------------
    Check the exact integral:
    INT.(1 to 3)[16 -x^2]dx
    = [16x -(x^3)/3] |(1 to 3)
    = [16(3) -(3^3)/3] -[16(1) -(1^3)/3]
    = [48 -9] -[16 -1/3]
    = 23.33 sq.units ------close to the approx. 23.25 sq.units.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Aug 2008
    Posts
    1

    x^3-6*x^2-x+30

    Quote Originally Posted by Misa-Campo View Post


    Can i please solution to this question , thanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Trapezoidal Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 13th 2010, 07:22 AM
  2. Simpsons rule,Trapezoidal rule,check please
    Posted in the Geometry Forum
    Replies: 0
    Last Post: February 16th 2010, 07:06 AM
  3. Replies: 0
    Last Post: October 20th 2008, 08:12 PM
  4. Simpsons Rule and Trapezoidal Rule
    Posted in the Calculus Forum
    Replies: 5
    Last Post: July 28th 2008, 04:50 AM
  5. Replies: 3
    Last Post: August 31st 2006, 10:08 AM

Search Tags


/mathhelpforum @mathhelpforum