Results 1 to 6 of 6

Math Help - Limit definition of definite integrals

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    242

    Limit definition of definite integrals

    In a practice quiz I'm working on for my calc class I'm asked to evaluate the given definite integral from 3 to 7 of (x^2-3x+5)dx using the limit definition of a definite integral. I know how to express an integral as a limit using the limit definition, and I know how to evaluate definite integrals usint areas, but I don't know how to EVALUATE a definite integral using the limit definition. What does this even mean exactly and how could I go about doing it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Abu-Khalil's Avatar
    Joined
    Oct 2008
    From
    Santiago
    Posts
    148
    You may define the partition x_0<x_1<\ldots<x_{n-1}<x_n, where x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right) so when n\to\infty, \Delta x_k \to 0.

    Finally, \int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5 and calculate that sum isn't a problem knowing \sum_{k=1}^nk=\frac{n(n+1)}{2} and \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2008
    Posts
    242
    thanks guys I'll check that link out for sure
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2008
    Posts
    242
    Quote Originally Posted by Abu-Khalil View Post
    You may define the partition x_0<x_1<\ldots<x_{n-1}<x_n, where x_k=3+\left(\frac{7-3}{n}\right)k\Rightarrow \Delta x_k=x_k-x_{k-1}=\left(\frac{7-3}{n}\right) so when n\to\infty, \Delta x_k \to 0.

    Finally, \int_3^7x^2-3x+5dx=\lim_{n\to\infty}\frac{7-3}{n}\sum_{k=1}^nx_k^2-3x_k+5 and calculate that sum isn't a problem knowing \sum_{k=1}^nk=\frac{n(n+1)}{2} and \sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}.
    I'm sorry that's a bit too jumbled for me I don't really understand any of what you said. What is x subscript k even representing here?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2008
    Posts
    242
    I keep getting wrong answers. I just found the answer by using the integral function on my calculator but I can't seem to get it. Could someone walk me through this step by step?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: December 5th 2011, 06:21 PM
  2. Replies: 3
    Last Post: May 31st 2011, 12:47 PM
  3. Replies: 4
    Last Post: April 13th 2011, 03:08 AM
  4. Limit of definite integrals
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: June 15th 2009, 06:35 AM
  5. definite integrals, upper limit variable
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: February 14th 2009, 04:47 AM

Search Tags


/mathhelpforum @mathhelpforum