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Evaluate the integral

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data

    integral from 0 to 7 (49-x^2)^(1/2)dx

    I just need help getting started; not sure how to find the anti-derivative of a function that is under a radical

    the way to do this is by using the FTC once I find the anti-D, right?




    3. The attempt at a solution
    I started messing around with it and the closest I could get was this:

    (2/3)(49-x^2)^(3/2)

    any help would be great
     
  2. jcsd
  3. Dec 10, 2007 #2

    rock.freak667

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    Try a trig substition like say, x=7sin[itex]\theta[/itex]
     
  4. Dec 10, 2007 #3

    quasar987

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    sin²+cos²=1 ==> 1-cos²=sin
     
  5. Dec 10, 2007 #4
    how did you guys know to do that?
     
  6. Dec 10, 2007 #5

    quasar987

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    When you see something of the form a-bx², this cries for a x=cos or x=sin substitution. Even more so when a-bx² is under a radical!
     
  7. Dec 10, 2007 #6
    Is there any other way to approach this problem? because we weren't taught that technique--we just learned the FTC last week
     
  8. Dec 10, 2007 #7

    quasar987

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    Well, the function is unbounded on (0,7) [look what happens near 7]... so it's not integrable automatically.
     
  9. Dec 10, 2007 #8
    trig substitute? we haven't learned any substitution methods yet... just the ftc...
     
  10. Dec 10, 2007 #9

    Avodyne

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    Maybe you're supposed to notice that this integral gives the area of a quarter-circle of radius 7.
     
  11. Dec 10, 2007 #10
    O_O

    wat
     
  12. Dec 10, 2007 #11

    quasar987

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    Nevermind that, I thought the function was 1/(...)^1/2
     
  13. Dec 10, 2007 #12

    quasar987

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    Maybe Avodyne has it.

    If you have a function y(x), what does the integral of y(x) from a to b represent? The area under the curve, I'm sure you know. Now what's the equation of a circle of radius 7? x²+y²=7². Isolate y and take the square root and keep the positive sign (this means, keep the part of the curve that is above the x axis), and you get (49-x²)^1/2. Therefor, the integral of (49-x²)^1/2 is the area of a quarter circle of radius 7.
     
  14. Dec 10, 2007 #13
    If you don't want to use trig subst. then one way is by expanding
    (49-x^2)^(1/2) using the binomial theorem.

    I dont know what you mean by FTC.

    The physical interpretation of this problem is exactly as said above, it is the area of a quarter-circle of radius 7 i.e. in the first quadrant.
     
  15. Dec 10, 2007 #14
    binomial theorem?

    ftc= fundamental theorem of calculus
     
  16. Dec 10, 2007 #15

    Avodyne

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    You don't want to know.
     
  17. Dec 10, 2007 #16
    since I haven't learned the trig sub. or the binomial theorem, either I recognize that is a half circle or I'm fu(ked?
     
  18. Dec 10, 2007 #17

    quasar987

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    Pretty much!
     
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