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Indefinite integral approximation technique

  • Thread starter cdotter
  • Start date
  • #1
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Homework Statement


[itex]\int_1^\infty \! \frac{(sin(x)+5)}{x^3} \, dx[/itex]
"Find two simpler integrals, one larger and one smaller."


Homework Equations





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?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,838
7
Can you find upper and lower bounds for sin(x)?
 
  • #3
305
0
Can you find upper and lower bounds for sin(x)?
sin(infinity) and sin(1)? I don't understand what you mean. :redface:
 
  • #4
Char. Limit
Gold Member
1,204
13
Yes, you can. Think of what y=sin(x) is restricted to...
 
  • #5
305
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Yes, you can. Think of what y=sin(x) is restricted to...
-1 and 1.
 
  • #6
Char. Limit
Gold Member
1,204
13
Exactly. So sin(x) is always less than or equal to 1.

You can use that.
 
  • #7
33,173
4,858
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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