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Integration Problem - given a function find a value

  1. Mar 27, 2009 #1
    Integration Problem -- given a function... find a value

    1. The problem statement, all variables and given/known data

    If f(3) = 7 and f'(x) = sin (1 + x^2) / (x^3 - 2x), then f(5) = ...


    2. Relevant equations

    We can use a calculator (TI-84+) to solve this problem. I'm having a brain fart ... we're given initial conditions, and we're given f'(x) a function that cannot be integrated with conventional methods.


    3. The attempt at a solution

    I'm thinking this has something to do with the manipulation of the integrand (and its respective limits).

    Usually, when I do these problems I integrate... find "C" and once "C" is a value, I plug in the value we're trying to find. But this function isn't easily integrable. What rules am I missing? I need hints not answers please, thank you!
     
  2. jcsd
  3. Mar 27, 2009 #2

    tiny-tim

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    Hi carlodelmundo! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Have you tried the obvious substitution … u = 1 + x2 ? :wink:
     
  4. Mar 27, 2009 #3
    Re: Integration Problem -- given a function... find a value

    Hello Tim!

    If we let u = 1 +x2 .. then du = 2x dx.

    This does us no good since the bottom term has degree 3. I know the bottom term can be factored but how can that help us?

    By the way... the equation is written as f'(x) = [ sin (1+x2) ] / (x3 - 2x). Thought I might clarify.
     
  5. Mar 27, 2009 #4

    tiny-tim

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    You don't seem to recognise the difference between a problem and an opportunity :wink:

    multiply top and bottom by x, and then the bottom will have degree 4 :smile:
     
  6. Mar 27, 2009 #5
    Re: Integration Problem -- given a function... find a value

    That is a very clever trick. Here is my work thus far:

    [tex]\int[/tex][tex]\frac{xsin (1 + x^2)}{x^4 - 2x^2}[/tex]

    [tex]\int[/tex][tex]\frac{xsin (1 + x^2)}{x^2(x^2-2)}[/tex]

    I let u = 1 + x2

    u - 1 = x2

    du = 2x dx

    The resulting integrand is:

    [tex]\frac{1}{2}[/tex][tex]\int[/tex][tex]\frac{sin (u) du}{(u-1)(u-3)}[/tex]

    From here I am completely lost. I'm thinking we need to use partial fractions to get rid of the ugly denominator? And sorry for the long response, I tried scanning only to use LaTex
     
  7. Mar 28, 2009 #6

    tiny-tim

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    Hi carlodelmundo! :smile:

    just got up :zzz: …
    Yes, partial fractions give you two integrals of the form ∫sinx/x …

    the integral of that is the "sine integral function", Si(x) … see http://en.wikipedia.org/wiki/Sine_integral

    I don't think there's any other way of doing it, except expanding sinx/x in powers of x, integrating each term, and then calculating to the required order of accuracy. :frown:
    General tip: when you're desperate :rolleyes:

    go back to the obvious! :biggrin:
     
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