Results 1 to 5 of 5

Math Help - Trapezium rule integration

  1. #1
    Member
    Joined
    Nov 2007
    Posts
    100

    Trapezium rule integration

    Find approximations to the value of:

    \int_{1}^5 \frac{1}{x^2} dx

    Using the trapezium rule with 4 intervals.

    Now I know that it's calculated using the formula:

    \frac{h}{2} (f(x_{0}) + f(x_{n})) + 2(f(x_{1}) + f(x_{2}) ... f(x_{n-1})) Where:

    n = interval

    h = \frac {b - a}{n}

    x_{k} = a + kh

    But the answer is not coming right :

    = 1/2 (1 + 1/25) + 2(1/4 + 1/9 + 1/16)

    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member McScruffy's Avatar
    Joined
    Jul 2009
    Posts
    87
    Awards
    1
    Quote Originally Posted by struck View Post
    Find approximations to the value of:

    \int_{1}^5 \frac{1}{x^2} dx

    Using the trapezium rule with 4 intervals.

    Now I know that it's calculated using the formula:

    \frac{h}{2} (f(x_{0}) + f(x_{n})) + 2(f(x_{1}) + f(x_{2}) ... f(x_{n-1})) Where:

    n = interval

    h = \frac {b - a}{n}

    x_{k} = a + kh

    But the answer is not coming right :

    = 1/2 (1 + 1/25) + 2(1/4 + 1/9 + 1/16)

    it should be \frac{1}{2}(f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4))

    which is \frac{1}{2}(f(1)+2f(2)+2f(3)+2f(4)+f(5))

    which is \frac{1}{2}\left(\frac{1}{1^2}+2\left(\frac{1}{2^2  }\right)+2\left(\frac{1}{3^2}\right)+2\left(\frac{  1}{4^2}\right)+\frac{1}{5^2}\right)
    Last edited by McScruffy; August 2nd 2009 at 10:08 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by struck View Post
    Find approximations to the value of:

    \int_{1}^5 \frac{1}{x^2} dx

    Using the trapezium rule with 4 intervals.

    Now I know that it's calculated using the formula:

    \frac{h}{2} (f(x_{0}) + f(x_{n})) + 2(f(x_{1}) + f(x_{2}) ... f(x_{n-1})) Where:

    n = interval

    h = \frac {b - a}{n}

    x_{k} = a + kh

    But the answer is not coming right :

    = 1/2 (1 + 1/25) + 2(1/4 + 1/9 + 1/16)

    \Delta{x}=\frac{5-1}{4}=1

    This implies that x_i=1+i\Delta{x}=1+i

    Then x_0=1,x_1=2,x_2=3,x_3=4,x_4=5

    So, we have By the TR

    \int_1^5\frac{1}{x^2}\approx\frac{1}{2}\Delta{x}[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    You beat me to it McScruffy! Thanx for the thanx, though!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member McScruffy's Avatar
    Joined
    Jul 2009
    Posts
    87
    Awards
    1
    Quote Originally Posted by VonNemo19 View Post
    You beat me to it McScruffy! Thanx for the thanx, though!
    No problem, I thought that your post explained how to get to where I started a bit better.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trapezium rule
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 9th 2011, 11:30 PM
  2. The trapezium rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: May 12th 2011, 06:27 AM
  3. Trapezium Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 14th 2010, 11:24 AM
  4. Trapezium Rule
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 9th 2010, 12:33 AM
  5. Integration/trapezium rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 24th 2010, 04:10 AM

Search Tags


/mathhelpforum @mathhelpforum