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Math Help - finding the area between two curves

  1. #1
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    finding the area between two curves

    Area between two curves:
    y=9-x^2
    y=x^2 - 7

    Show that the area is (128√2)/3

    So...
    y=9-x^2
    y'=-2x
    I(x) = 9x - (1/3)x^3

    y=x^2 - 7
    y'=2x
    I(x) = (1/3)x^3 - 7x

    9 - x^2 = x^2 - 7
    x = √8

    Correct so far? What next? How can I make it to find the correct area? Thanks!
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  2. #2
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    Quote Originally Posted by rubbersoul923 View Post
    Area between two curves:
    y=9-x^2
    y=x^2 - 7

    Show that the area is (128√2)/3

    So...
    y=9-x^2
    y'=-2x
    I(x) = 9x - (1/3)x^3

    y=x^2 - 7
    y'=2x
    I(x) = (1/3)x^3 - 7x

    9 - x^2 = x^2 - 7
    x = √8

    Correct so far? What next? How can I make it to find the correct area? Thanks!
    x = =/- root 8
    These are then the limits for the integration.
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  3. #3
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    So then the -√8 and √8...

    I plug them into each integral
    I(x) = 9x - (1/3)x^3
    I(√8) = 9(√8) - (1/3)(√8)^3 = 17.9
    I(-√8) = 9(-√8) - (1/3)(-√8)^3 = -17.9

    I(x) = (1/3)x^3 - 7x
    I(√8) = (1/3)(√8)^3 - 7(√8) = -12.2
    I(-√8) = (1/3)(-√8)^3 - 7(-√8) = 12.2


    How do I arrive at the area of 128√2 / 3 ?

    Area = b-a
    Area = which one is b and which is a?
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  4. #4
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    Quote Originally Posted by rubbersoul923 View Post
    So then the -√8 and √8...

    I plug them into each integral
    I(x) = 9x - (1/3)x^3
    I(√8) = 9(√8) - (1/3)(√8)^3 = 17.9
    I(-√8) = 9(-√8) - (1/3)(-√8)^3 = -17.9

    I(x) = (1/3)x^3 - 7x
    I(√8) = (1/3)(√8)^3 - 7(√8) = -12.2
    I(-√8) = (1/3)(-√8)^3 - 7(-√8) = 12.2


    How do I arrive at the area of 128√2 / 3 ?

    Area = b-a
    Area = which one is b and which is a?
    Draw a quick sketch. y = 9-x^2 is an "upside-down parabola" with y -int 9 and axis of symmetry the y-axis. Also draw y = x^2 -7 (you should know what it looks like). So you want to calculate the integral from -rt8 to +rt8 of the function ((9-x^2) - (x^2 -7))
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