1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral Problem help!

  1. May 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate the integral...

    int(0 to 2) of 40/(x^2 + 4)dx

    a: Your answer should be in the form ` kpi `, where ` k ` is an integer. What is the value of ` k `?

    b: Now, lets evaluate the same integral using power series. First, find the power series for the function ` f(x) = (40)/(x^2+4) `. Then, integrate it from 0 to 2, and call it S. S should be an infinite series.

    What are the first few terms of S ?

    c: The answers to part (a) and (b) are equal (why?). Hence, if you divide your infinite series from (b) by ` k ` (the answer to (a)), you have found an estimate for the value of ` pi ` in terms of an infinite series. Approximate the value of ` pi ` by the first 5 terms.

    d: What is the upper bound for your error of your estimate if you use the first 10 terms? (Use the alternating series estimation.)

    3. The attempt at a solution

    k is 5, this is correct, I know how to get k...

    the first term of the series is 20, I know how to get that, but then the rest of my series goes wacky.

    Here's my equation for the series

    the sum of (-1)^n * ((2)^(2n + 1)/(10^(n - 1) * (2n + 1))

    I can show how I got that in case anyone wants, but I think it's fairly obvious.

    And at the end of that, using exact integral notation, there'd be the same equation but instead of that first 2 there'd be a 0, but that simplifies to 0 so it doesn't matter.

    When I use n = 0, I get 20 like I'm supposed to, but when I use n = 1, i get -2.6667...answer is -6.6667

    my next term is .64, should be getting 4...then -.182857, should get a -2.85714285714...then .0568888889, should get a 2.22222222222...

    For c and d of course I'm gonna be off, but I got 3.5694 and 3.64722E-6 respectively. The correct answers are 3.33968253968 and 0.190476190476.

  2. jcsd
  3. May 27, 2008 #2
    OK. So, first of all, your series is close, but not correct. Since [tex]\frac{40}{x^2+4}[/tex] is the same as [tex]\frac{10}{1+\frac{x^2}{4}}[/tex], for all x in the interval of convergence,

    [tex]{\int_0^2 f(x) dx} = 10 {\int_0^2 {\sum_n \left(\frac{-1}{4}\right)^n x^{2n} } dx} = 10{\sum_n \left(\frac{-1}{4}\right)^n \frac{2^{2n+1} }{2n+1}}[/tex]

    This should fix your problem. :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook