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Finding the area between 3 curves

  1. Oct 30, 2009 #1
    fx=3x^3-3x, gx=3x, and hx=9-x. Find the area

    I kown top - bottom and right - left. but in here i am not sure what to do and what the boundaries are. can some one show me the work how to do this problem??? i am kinda confuse how to do this kind of problem with 3 curves. THANK YOU!!!
     
  2. jcsd
  3. Oct 30, 2009 #2
    Did you draw a picture? The region looks like a triangle, but the side formed by 3x^3-3x is curved. I would start by fining the verticies of the triangle... or the points where these lines intersect.

    So we need to solve 3 systems of equations. Once you have done that you can divide the region in to a true triangle and a curved slice. Find the areas of each separately.
     
  4. Oct 30, 2009 #3
    sorry still don't know what to do. my book doesn't have any examples like this question
     
  5. Oct 30, 2009 #4
    Look at this image:

    4059633926_f5e8fc5039_m.jpg

    First find the points where the graph intersects to form the triangle by solving these three systems of equations:

    System 1
    [tex]y=3x^3-3x[/tex]
    [tex]y=3x[/tex]

    A(x,y) =

    System 2
    [tex]y=3x^3-3x[/tex]
    [tex]y=9-x[/tex]

    B(x,y) =

    System 3
    [tex]y=9-x[/tex]
    [tex]y=3x[/tex]

    C(x,y) =

    Now you can break it in to two differences of integrals.

    One from A to B and another from B to C.
     
  6. Oct 31, 2009 #5
    ok i graphed and i found the 3 points. A:(1.4142,4.2426) B:(1.5958,7.4042) C:(2.25,6.75)
    so when u say break it into 2 parts you mean (∫1.5958 on top, 1.4142 bottom (9-x)-(3x) dx)???? am i heading to the right direction??? if i am i don't know how to get the other part
     
  7. Oct 31, 2009 #6
    The first integral will be:

    [tex]\int_{x_1}^{x_2}(x^3-3x) -\int_{x_1}^{x_2}3x[/tex]

    Where x1 and x2 are the x values from the points A and B. then add that to the 2nd integral:

    [tex]\int_{x_2}^{x_3}(9-x) -\int_{x_2}^{x_3}3x[/tex]

    Where x2 and x3 are the x values from the points B and C.

    Don't just take my word for it! Make certian you understand *why* --think about the region each integral represents, shade in the graph if needed.

    The values you found look reasonable, but it looks like you used a graphing calculator? If I were teaching this course I'd want an exact value in radicals. Just check that your prof. is OK with aprox. values.
     
  8. Oct 31, 2009 #7
    HA. i see. and to get the exact radical you just set 9-x=3x^3-3x to get B and so on right? thank you so much btw
     
  9. Oct 31, 2009 #8
    Yup. You'll get 3 solutions for that since the graphs intersect 3 times. Just pick the one with the largest value...
     
  10. Oct 31, 2009 #9
    oops wait a minute why do i get a negative value when i do ∫ x^3-3x? and the bonds are 1.5958 and 1.4142 right? shoulden't they all be positive?

    nvm my mistake
     
    Last edited: Oct 31, 2009
  11. Oct 31, 2009 #10
    They should all be positive.
     
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