Results 1 to 3 of 3

Math Help - [SOLVED] easy limits area under the function

  1. #1
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230

    Angry [SOLVED] easy limits area under the function

    The value of the limit lim n--> inf E (n above and i=1 below) pi/8n tan(ipi/24n) is equal to the area below the graph of a function f(x) on an interval [A,B]. Find f, A, and B. (Do not evaluate the limit.) for f i get tan(x) for a i get -pi/12 for b i get .1308996939 i used delta x = (b-a)/n i also used xi = a + (delta x)(i) so i know a + pi/8n = ipi/24n solved for a then (b-a)/n = pi/8n solved for b my webwork is saying its wrong???? help please. i dont know what else to do
    Last edited by mybrohshi5; November 25th 2009 at 08:18 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by mybrohshi5 View Post
    The value of the limit lim n--> inf E (n above and i=1 below) pi/8n tan(ipi/24n) is equal to the area below the graph of a function f(x) on an interval [A,B]. Find f, A, and B. (Do not evaluate the limit.) for f i get tan(x) for a i get -pi/12 for b i get .1308996939 i used delta x = (b-a)/n i also used xi = a + (delta x)(i) so i know a + pi/8n = ipi/24n solved for a then (b-a)/n = pi/8n solved for b my webwork is saying its wrong???? help please. i dont know what else to do
    \lim_{n\to\infty}\sum_{i=1}^n\frac{\pi}{8n}\tan\le  ft(\frac{i\pi}{24n}\right)=\lim_{n\to\infty}\sum_{  i=1}^n\frac{\pi}{8n}\tan\left(\frac{i\frac{\pi}{8n  }}{3}\right)

    From this you can see that the function is \tan\left(\frac{x}{3}\right), and your interval length is \frac{\pi}{8}.

    However, since you have no constant being added to the argument of tangent, your interval is from 0 to something. Since the length of the interval is \frac{\pi}{8}, this implies that the upper bound is 0+\frac{\pi}{8}=\frac{\pi}{8}.

    \lim_{n\to\infty}\sum_{i=1}^n\frac{\pi}{8n}\tan\le  ft(\frac{i\pi}{24n}\right)=\int_0^{\pi/8}\tan\left(\frac{x}{3}\right)\,dx=-3\ln\left(\cos\left(\frac{\pi}{24}\right)\right)\a  pprox0.258

    PS:

    Quote Originally Posted by mybrohshi5 View Post
    for b i get .1308996939
    This value is \frac{\pi}{24}, in case you were wondering.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member mybrohshi5's Avatar
    Joined
    Sep 2009
    Posts
    230
    Great explanation. That makes way more sense then how i figured it out a little bit ago. thank you very much =)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Some easy ->infinity limits I can't solve
    Posted in the Calculus Forum
    Replies: 7
    Last Post: August 13th 2010, 02:29 AM
  2. Replies: 2
    Last Post: September 27th 2009, 12:51 PM
  3. Replies: 1
    Last Post: February 16th 2009, 03:43 PM
  4. [SOLVED] Express the surface area as a function
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: February 3rd 2009, 12:01 PM
  5. [SOLVED] and now something easy
    Posted in the Math Challenge Problems Forum
    Replies: 5
    Last Post: July 20th 2006, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum