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Area under curve questions

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the areas enclosed by the following curves and straight lines:
    c) y= (1/x2) -1 , y= -1 , x=1/2, and x=2

    b) y = x3-1, the axes and y = 26






    2. The attempt at a solution
    Okay I sketched the curve and to me it looks like the curve occupies no area at y=-1 it's undefined there.

    so isn't the area occupied from x = 1/2 to x =2

    so ∫1/x2) -1 . dx
    y= [(1/x) - x ] + c

    ( (1/2) - 2) - ( 2 - (1/2)) = -3
    but the answer in my book says 3/2

    and for the 2nd question
    The points at which these curves intersect are (3,26)
    ∫x3 -1 . dx
    y = (x4/4) - x = 17 1/4 ( x = 0 to 3)

    ∫26.dx = 26x = 78 sq units( from x 0 to 3)

    78- 17/4 = 60 3/4 but in the back of my book it says 60 exactly, and I don't see what I did wrong..

    Help is greatly appreciated~!!
     
  2. jcsd
  3. Feb 5, 2013 #2

    Dick

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    The area between two curves f(x)>g(x) is the integral of f(x)-g(x). In the first case, f(x) is 1/x^2-1 and g(x) is (-1).
     
  4. Feb 5, 2013 #3
    So you mean just to (1/x)-x -x? and that's the area?
    But what do I put in for the values of x.. 1/2 and 2?
     
  5. Feb 5, 2013 #4

    Dick

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    The function to integrate is ((1/x^2)-1)-(-1). Yes, put the x limits to 1/2 and 2. Your previous attempt simply ignored the y=(-1) for reasons that are somewhat obscure to me. It's true that the two curves don't cross. But you can't ignore the lower curve.
     
  6. Feb 5, 2013 #5
    I thought that it had no area at y=-1 because when y=-1 the function is undefined .

    So do the same thing for the question before?

    >.> sorry it's not undefined I don't know what I was thinking about.................
     
  7. Feb 5, 2013 #6

    Dick

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    There is no point where 1/x^2-1 equals (-1). Correct. That just means the two curves don't intersect each other. Doesn't mean you can ignore one.
     
  8. Feb 5, 2013 #7
    When I use 1/x I keep getting -3/2
     
  9. Feb 5, 2013 #8

    Dick

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    Show how you got that. Hint: the integral of 1/x^2 is not 1/x.
     
  10. Feb 5, 2013 #9
    =.= my god .... didn't see it's -1/x forgot to divide by -1
     
  11. Feb 5, 2013 #10
    but for the 2nd question I can't get the exact 60 I get 60 3/4
    Isn't the integral [ (x^4/4) -27x] ?
     
  12. Feb 5, 2013 #11

    Dick

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    You really need to draw a sketch of the region you are integrating over. You are working blind here. You don't integrate over that function for the whole interval [0,3]. At some point x^3-1 will cross the x-axis. That's supposed to be one of your boundaries. You need to split the integral up.
     
  13. Feb 5, 2013 #12
    Okay so I integrate the region x=0 to x=1 and then x= 1 to x=3

    then subtract them?
     
  14. Feb 5, 2013 #13

    Dick

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    I would say add them. And what you are integrating in each region is different. What two different forms are they? Why do you say subtract them?
     
  15. Feb 5, 2013 #14
    Oh because I was thinking about what you said area between curves is
    f(x) - g(x) but now when I look at my sketch that wouldn't make sense.
     
  16. Feb 5, 2013 #15

    Dick

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    The graph will tell you what to do.
     
  17. Feb 5, 2013 #16
    I just don't understand! The integral is ((x^4/4) -27x) right? when I integrate from x=0 to x=1 I get -26 3/4

    and when I do it for x=1 to x=3 I get -34.. I can't get the +60 !
     
  18. Feb 5, 2013 #17

    Dick

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    What are the boundaries between x=0 and x=1? Hint: Neither of them is x^3-1. Look at your graph! The upper boundary is 26 and the lower boundary is the x-axis!!
     
  19. Feb 5, 2013 #18
    I think I got it now... the integral for x=0 and x=1 is 26x so it's 26..

    then for the other side of the graph it's (x^4/4 -x) for x=3 and x=1 and that's 18
    then I find the one for 26x for x = 3 and x= 1 and I get 52

    Then I subtract the areas and get 34 then I add it to the area bounded by the 26
    and I got 60.
     
  20. Feb 5, 2013 #19

    Dick

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    Yes, x=0 to x=1. It's 26. From x=1 to x=3 it's the integral of 26-(x^3-1) which gives 34. Then you just add them. But your way works too.
     
  21. Feb 5, 2013 #20
    Umm 1 question I know this might be common sense.. but
    on these graphs, the one that looks like it covers more area is the one that I will subtract whatever the function? Like y=26 looks like it covers more than x^3-1.. but maybe it looks that way cause of my sketch sometimes I can't trust my sketch that much cause my work is very untidy.
     
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