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Homework Help: Area between curves

  1. Jan 5, 2004 #1
    I need to find the area between y=x and y=x^2
    So this is what I did:
    [tex] A = \int (x^2-x) \,dx [/tex]
    Then I found the limits of integration x=0 and x=1 because thats where the two graphs intersect
    [tex] A = \int^1_0 (x^2-x) \,dx [/tex]
    I ended up with an answer of -1/6
    What did I do wrong?
     
  2. jcsd
  3. Jan 5, 2004 #2

    ShawnD

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    intersect points:
    [tex]x^2 = x[/tex]

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

    [tex](x)(x-1) = 0[/tex]

    [tex]x = 1, x = 0[/tex]



    The upper limit is the line [tex]y = x[/tex], the lower limit is [tex]y = x^2[/tex]

    [tex]A = \int^1_0 x \,dx - \int^1_0 x^2 \,dx [/tex]

    [tex]A = \frac{x^2}{2} |^1_0 - \frac{x^3}{3} |^1_0[/tex]

    [tex]A = \frac{1^2}{2} - \frac{1^3}{3}[/tex]

    [tex]A = \frac{1}{2} - \frac{1}{3}[/tex]

    [tex]A = \frac{1}{6}[/tex]



    Your answer seems fine to me.
     
  4. Jan 5, 2004 #3

    Hurkyl

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    Which one's bigger?
     
  5. Jan 5, 2004 #4
    oh...... I get it
     
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