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Indefinite integral approximation technique

  1. Mar 19, 2010 #1
    1. The problem statement, all variables and given/known data
    [itex]\int_1^\infty \! \frac{(sin(x)+5)}{x^3} \, dx[/itex]
    "Find two simpler integrals, one larger and one smaller."


    2. Relevant equations



    3. The attempt at a solution

    How could I make this a simpler (ie, solvable) integral? It's been straight forward with other integrals like this:

    [itex]\int_1^\infty \! \frac{1}{\sqrt{x^4+10}} \, dx \rightarrow \int_1^\infty \! \frac{1}{\sqrt{x^4}} \, dx[/itex]

    Doing this would give an overestimate since the denominator is smaller. But how would I do it for the integral in question?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 19, 2010 #2
    Can you find upper and lower bounds for sin(x)?
     
  4. Mar 19, 2010 #3
    sin(infinity) and sin(1)? I don't understand what you mean. :redface:
     
  5. Mar 19, 2010 #4

    Char. Limit

    User Avatar
    Gold Member

    Yes, you can. Think of what y=sin(x) is restricted to...
     
  6. Mar 19, 2010 #5
    -1 and 1.
     
  7. Mar 19, 2010 #6

    Char. Limit

    User Avatar
    Gold Member

    Exactly. So sin(x) is always less than or equal to 1.

    You can use that.
     
  8. Mar 19, 2010 #7

    Mark44

    Staff: Mentor

    And so ? <= sin(x) + 5 <= ?

    Filling in the question marks should make two definitie integrals (not indefinite integrals) that are a lot easier to evaluate.
     
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