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Area between two curves

  1. Aug 27, 2008 #1
    I have three questions:

    1. The problem statement, all variables and given/known data

    Find the areas of the regions whose boundaries are given.

    2. Relevant equations

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

    3. The attempt at a solution

    x=-2, x=2
    I got -10.67 but I know this can't be true because you can't have a negative area.



    1. The problem statement, all variables and given/known data

    Find the areas of the regions whose boundaries are given.

    2. Relevant equations

    [tex]y^2=x[/tex]
    [tex]x+y=2[/tex]

    3. The attempt at a solution

    y=1, y=-2
    I got -4.5, but again that can't be right because it's negative :(


    1. The problem statement, all variables and given/known data

    Find the areas of the regions whose boundaries are given.

    2. Relevant equations

    [tex]y=x^3-2x^2-3x[/tex]
    [tex]y=0[/tex]

    3. The attempt at a solution

    I got to x=0, x=-1, and x=3 but I don't know where to go from here.

    Thanks for any help! :)
     
  2. jcsd
  3. Aug 27, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    Let's just start with the first one. How did you get x=-2, x=2 or was that given? The curves y=x^3-3 and y=1 don't enclose any bounded region.
     
  4. Aug 27, 2008 #3
    many apologies... it should have been x^2-3
     
  5. Aug 28, 2008 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    If you get a negative area then you have the two functions in the wrong order. For x between -2 and 2, the graph of x2- 3 is below the graph of y= 1. You should be integrating [itex]\int [1- (x^2-3)]dx= \int (4- x^2)dx[/itex]. That, integrated between -2 and 2, is positive.
     
  6. Aug 28, 2008 #5
    And for the last problem, you would have two different enclosed areas. One would be between -1 and 0 and the other between 0 and 3.
    This means you would have to set up 2 different equations and find the sum of the areas.
    HINT: The equations are very similiar just siwtched in order.
     
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