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Another area under curve problem

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Sketch roughly the curve y = x^2(3-x) between x=-1 and x=4. Calculate the area bounded by the curve and the x-axis


    . The attempt at a solution

    I tried to find the area from x=-1 to x=4 I got 1 1/4
    answer in the back of my textbook is 6 3/4

    When i find the area from x= 0 to x= 3 I get 6 3/4.

    What should I do?
     
  2. jcsd
  3. Feb 7, 2013 #2

    SammyS

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    So, what exactly was it that you integrated?
     
  4. Feb 7, 2013 #3
    I integrated x^2(3-x) to 3x^2-(x^4/4) for x= -1 to x=4 and I got 1 1/4

    then I tried x= 0 to x=3 and i got 6 3/4 the answer in the back of the book but I don't understand why i got it, I got the answer for the wrong reason =/
     
  5. Feb 7, 2013 #4

    SammyS

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    Is that the exact wording of the problem.

    It appears that the intent is for you to find the area of the region enclosed by y = x2(3-x) and the x-axis.

    What were the x-intercepts on your graph?
     
  6. Feb 7, 2013 #5
    -1 and 3 and yes I typed the exact question word for word1
     
  7. Feb 7, 2013 #6

    SammyS

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    Only the region between x=0 and x=3 is enclosed by y = x2(3-x) and the x-axis.
     
  8. Feb 7, 2013 #7
    so therefore the answer is 6 3/4

    oh wow sorry -1 is not an intercept.. it's 0(twice) and 3
    so basically the x=-1 and x=4 are things to trick me?
     
  9. Feb 7, 2013 #8

    SteamKing

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    Check your indefinite integration. It appears the curve is y = x^2*(3-x), which is
    y = 3x^2 - x^3 when expanded
     
  10. Feb 7, 2013 #9

    HallsofIvy

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    The graph is below the x-axis for x from x=3 to x= 4. The area is given by
    [tex]\int_{-1}^3 3x^2- x^3 dx- \int_3^4 3x^2- x^3 dx[/tex]
     

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  11. Feb 7, 2013 #10
    Sorry but I'm confused that would make it negative and that would mean the area is under the x-axis. But my book implies it's above.
     
  12. Feb 7, 2013 #11
    According to how you put it, there was no mention of the area being above the x-axis. Usually if this is the case, the question would explicitly say to find the region above the x-axis rather than the area bounded by a curve and the x-axis.
     
  13. Feb 7, 2013 #12
    Does this help:
    [tex]\int_{-1}^3 3x^{2}-x^{3}dx-\int_3^4 3x^{2}-x^{3}dx[/tex]

    In order to get the AREA of the curve, you have to add the 2 parts of it (where it's positive and negative). If you just do the integral from -1 to 5 you end up with 5/4, which isn't the answer you need.

    And I don't get the answer your book does, so either you have the problem written wrong or the books answer is wrong (or I'm wrong lol)
     
    Last edited: Feb 7, 2013
  14. Feb 7, 2013 #13

    SammyS

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    Look at the problem statement again.

    The only mention of x=-1 and x=4 are in regards to sketching the graph.

    The graph of y = x2(3-x) intersects the x-axis at x=0 and x=3.

    Therefore, I think you were expected to find the area below y = x2(3-x) and above y = 0, between x=0 and x=3 .

    i.e., you were to find [itex]\displaystyle \ \ \int_{0}^{3} x^2(3-x)\,dx\ .[/itex]
     
  15. Feb 7, 2013 #14
    I agree with SammyS because the other questions I couldn't do, I got the answers when I integrated for the parts of the graph cutting the x.
     
  16. Feb 8, 2013 #15
    You should consider learning from another textbook. The area based on the wording should be what HallsofIvey has. The answer that SammyS gave is only involving the 1st quadrant! You can't simply say the area bounded by the curve and the x-axis ignores the 2nd and 4th quadrants because that is very relevant!
     
  17. Feb 8, 2013 #16

    SammyS

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    Of course I disagree with the statement regarding the area. I did initially interpret the question the same way that HallsfoIvy did. The fact that the integral from x=0 to x=3 gave the area matching the answer key made me see if I could interpret the question in a different light.

    As for the textbook, it does appear at first glance that the problem could have (and should have) been worded more clearly. However, without examining the textbook in more detail -- particularly looking at the way the material was presented and looking at the examples given -- I can't make a such judgement regarding the textbook -- good or bad.


    Had the problem been just a bit different, as in the following, I think there would have been no confusion. Try this out. (I've just shifted everything down by 2 units.)
    Roughly sketch the curve y = x^2(3-x)-2 . Be sure the graph extends at least from x=-1 to x=4. Calculate the area bounded by the curve and the line, y=-2 . (It may be helpful to notice that x^2(3-x)-2 = (1-x)(-2-2 x+x^2) . )​
     
  18. Feb 9, 2013 #17
    Is the answer 23.3?
     
  19. Feb 9, 2013 #18

    SammyS

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    If I didn't mess up the problem, the answer is 6 ¾ .
     
  20. Feb 9, 2013 #19
    For the one you posted?
     
  21. Feb 9, 2013 #20
    Maybe I messed up the math cause the roots for the one you posted were weird
     
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