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A little problem

  1. Dec 5, 2004 #1
    Easy problem... but urgent!

    y = (x^5 + 4x^4)^(1/2)

    They want me to find the area of this loop, and the boundaries are -4 and 0, but when I integrate it and plug in the -4, I get 0, which is clearly not the case. If you can help, I would love it.
     
    Last edited: Dec 5, 2004
  2. jcsd
  3. Dec 5, 2004 #2

    Tide

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    I think you left out the second limit on the integration!
     
  4. Dec 5, 2004 #3

    WaR

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    Can you post your integrating steps?
    That integration is not equal to [tex]0[/tex]
    But
    [tex]\int_{-4}^{0} \sqrt{x^{5}+4x^{4}} dx = \frac{2048}{105}[/tex]
     
  5. Dec 5, 2004 #4
    I did:

    Integration of (x^5 +4x^4)^(1/2) is 2/3 (x^5 +4x^4)^(3/2)

    So 2/3(0 + 0)^(3/2) = 0

    0 - [2/3 (-1024 + 1024)^(3/2) = 0

    It doesn't work.
     
  6. Dec 5, 2004 #5

    Tide

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    No, you integrated incorrectly!
     
  7. Dec 5, 2004 #6
    ok, the answer given in the back of the book is 4096/105. I know that even if that answer is not correct, it is more correct than mine because the area of that loop in the graph is clearly not 0.
     
  8. Dec 5, 2004 #7
    ok, I think the book is wrong because the calculator said that it was 19.50476233. I still don't know how this works though because.... I still get 0.
     
  9. Dec 5, 2004 #8

    Pyrrhus

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    Little hint

    [tex] \int_{-4}^{0} \sqrt{x^{5}+4x^{4}} dx [/tex]

    [tex] \int_{-4}^{0} x^{2} \sqrt{x+4} dx [/tex]
     
  10. Dec 5, 2004 #9
    that's how the original problem was set up. I distributed, thinking it would be easier. In any case, I get 0. Thanks
     
  11. Dec 5, 2004 #10

    Pyrrhus

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    Post the original problem as the book states it.
     
  12. Dec 5, 2004 #11
    Sketch the graph y^2 = x^4(x+4) and find the area enclosed by the loop.
     
  13. Dec 5, 2004 #12

    Tide

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    That is distinctly different from the problem you first posed - but your integration is still incorrect! :-)
     
  14. Dec 5, 2004 #13
    the boundaries are still -4 and 0, and my answer still comes out to be 0. What is correct? I guess I am not understanding. Please help!!
     
  15. Dec 5, 2004 #14

    Tide

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    [tex]\int x^2 \sqrt {x+4} dx = \frac {2 (x+4)^{3/2} (15 x^2 -48 x + 128)}{105}[/tex]
     
  16. Dec 5, 2004 #15
    ok so 2/3(x+4)^3/2, I got that part, but how did you get the other part. I am not familiar with integrating something undistributed unless it follows the f(x) * f'(x) rule. Sorry for all the questions. Thank you so much!!
     
  17. Dec 5, 2004 #16
    Use cyclovenom's hint
     
  18. Dec 5, 2004 #17
    Now I am totally lost.... I know I have to integrate only the top half of this loop and multiply it by 2 to find the area or it will equal 0. I just do not know how to set up the problem to make it only the top half of the loop.
     
    Last edited: Dec 5, 2004
  19. Dec 5, 2004 #18

    Tide

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    You can integrate by parts. You may get a slightly different form than I did because I did some simplification.
     
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