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Question about Integration

  1. Feb 21, 2007 #1
    I was wondering if I did this correctly:

    Sketch the region enclosed by y = x and y^2=x+2 and then determine its area.

    I got the area enclosed to be x = 0, x = 2.

    Then I used Integral from 0 to 2 (x+2)^(1/2) - (x) = [(2/3)(x+2)^(3/2) - (x^2/2)] evaluated at 2 and 0 which = 10/3
     
  2. jcsd
  3. Feb 21, 2007 #2

    hage567

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    Are you sure? Your method looks fine but I think you may have made a mistake when you put x=0 in your result. (2/3)(0+2)^(3/2) is not zero.
     
  4. Feb 21, 2007 #3
    oo your right.. goddamn.. lol.. so does this look right

    10/3 - (4*(2)^(1/2))/3

    I just wasnt sure if I went from the right points (0,2) because the graph y = (x+2)^(1/2) ends at x = -2.
     
  5. Feb 21, 2007 #4

    hage567

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    3, you mean? not that it's going to matter...

    I just graphed it and I see what you mean. Yeah, I guess you should take it from x=-2, since there is nothing in your question about x=0.
     
  6. Feb 21, 2007 #5
    alright so integrate from 2 to -2 which i would get:

    16/3?
     
  7. Feb 21, 2007 #6

    hage567

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    I agree with that answer.
     
  8. Feb 21, 2007 #7
    alright man.. Thanks..
     
  9. Feb 21, 2007 #8

    hage567

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    You're welcome. :smile:
     
  10. Feb 21, 2007 #9
    one more question:

    I need total distance of a particle traveling at v(t) = t^3 - 3t for [0,5]

    t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 5 and x = 0

    = 475/4?? Did i do this correctly for the TOTAL distance
     
    Last edited: Feb 21, 2007
  11. Feb 21, 2007 #10
    Anyone????
     
  12. Feb 21, 2007 #11

    hage567

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    Hmmm, yeah I would be careful with that. If you put in just x=0 and x=5 you would get the displacement, which is not necessarily the total distance. So you have to know which they are really asking for. Try adding up the distance between t=0 to t=1, then t=1 to t=2 up to t=5 to get a better idea of the total distance.
     
  13. Feb 21, 2007 #12
    so

    t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 1 and x = 0
    +
    t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 2 and x = 1
    +
    t^3 - 3t = (t^4/4) - (3/2)t^2 at x = 3 and x = 2

    etc?? Like that??
     
  14. Feb 21, 2007 #13

    hage567

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    Yeah, I would try that.
     
  15. Feb 21, 2007 #14
    All added together i get 481/4??

    is the any definite way to know the right answer..
     
  16. Feb 21, 2007 #15

    hage567

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    I can't think of one of off the top of my head. Is there someone you can clarify the question with?
     
  17. Feb 21, 2007 #16
    not really.. but i do have another question for ya if your up for it..


    Yes I know i have a lot, but I am being tested quite soon and I need to know my ****
     
  18. Feb 21, 2007 #17

    hage567

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    I'll give it a try. You also might see more action if you post your questions in the Homework Forums.
     
  19. Feb 22, 2007 #18
    Awesome.. heres one..

    find Integral of (t^2)/(t-1)^(1/2) dt

    I know we need to use the substitution rule.. but I am not sure where to begin:

    u = t^2, du = 2t dt, dt = du/2t??
     
  20. Feb 22, 2007 #19

    arildno

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    Dearly Missed

    Integration by parts works nicely here.
     
  21. Feb 22, 2007 #20
    i know but Iam unsure of where to begin..
     
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