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Area bounded by curves-Integral

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the area bounded by the curves, y= √x, y= (5-x)/4, and y= (3x-8)/2

    2. Relevant equations



    3. The attempt at a solution

    I found the intersection between each of the three curves to each other. Not sure what exactly the area bounded is. Is it the small triangular area underneath all three with (5,0) as one vertex, and (3,1/2) as another or?

    (3, 1/2)
    (16/9, -4/3)
    (1,1)

    ^These are what I think the three intersection points are.

    ∫((5-x)/4) - √x with b= 1 but I'm not sure what the lower limit is. I'm really stuck. Guidance please?
     
  2. jcsd
  3. May 28, 2013 #2

    Dick

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    (16/9, -4/3) can't be an intersection point with y=sqrt(x). -4/3 is negative. Try that one again.
     
  4. May 28, 2013 #3
    Oops! It's 4/3 (dumb mistake).
     
  5. May 28, 2013 #4

    Dick

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    No, -4/3 is the correct y value if x=16/9. So that's not an intersection. But there's another x value that is.
     
  6. May 28, 2013 #5
    So for my intersection point I get:

    2* sqrt(x) = (3x-8)^2
    4x= 9x^2 - 48x + 64
    9x^2 - 52x + 64 =0
    x= 4 or x= 16/9
    x= 4 therefore y= 2

    Is (4,2) the right one?
     
  7. May 28, 2013 #6

    Dick

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    Yes, it is. So you want the area of the sort of triangular region between those three points.
     
  8. May 28, 2013 #7
    So the three points are (1,1) (3, 1/2) and (4,2).

    From 1 to 3, (5-x)/4 is greater than sqrt(x). So I would find the integral of ((5-x)/4 - sqrt(x)) on the interval (1,3)? And the other interval would be (3,4) and the integral on this interval would be (3x-8/2) - (5-x/4)?

    Am I doing this right? And I know there should be a third integral but I can't figure it out..
     
  9. May 28, 2013 #8

    Dick

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    No, there's no third integral. But you should always sketch a graph when you are doing a problem like this. You ideas about what is greater than what are wrong. But you've got the right idea how to set the integrals up.
     
  10. May 28, 2013 #9
    I did sketch them, but I guess I need to pay more attention. :eek:

    So 5-x/4 is NOT greater than sqrt(x) on the interval from 1 to 3?
    And 3x-8/2 is similarly not greater than 5-x/4 on (3,4)?

    I don't understand how that holds true.. Please explain? :/
     
  11. May 28, 2013 #10

    Dick

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    I'd be interested in seeing what your sketch looks like. 2 is in (1,3). If you put x=2 into (5-x)/4 you get 3/4. If you put it into sqrt(x) you get sqrt(2). Which is greater?
     
  12. May 28, 2013 #11
    sqrt(2) is greater. I realized that I didn't plug in the correct values! But using the same logic, 3x-8/2 is still greater than 5-x/4 if I'm not mistaken?

    Therefore it would be sqrt(x) - ((5-x)/(4)) with a= 1 and b= 3.
    And 3x-8/2 - ((5-x)/(4)) with a= 3 and b= 4?
     
  13. May 28, 2013 #12

    Dick

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    In my graph it looks like the area you want is bounded above by sqrt(x) and below by ((5-x)/4) on the interval (1,3) and above by sqrt(x) and below by (3x-8)/2 on (3,4). The only regions I see that bounded above by one line and below by the other aren't part of the region you want to find the area of.
     
  14. May 28, 2013 #13
    I graphed it on a calculator, and I see what you are referring to. My calculation for the total area is 23/12. Is this correct? Thanks.
     
  15. May 28, 2013 #14

    Dick

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    Yes, that's correct.
     
  16. May 28, 2013 #15
    Thank you so much. Now I leave with a much better understanding of this concept than I arrived with! :)
     
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