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Math Help - finding the area using integration?

  1. #1
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    finding the area using integration?

    y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance
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    Quote Originally Posted by takermaster35 View Post
    y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance
    2 \int_0^1 x \sqrt{x} (1-x)^3 \ dx.
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  3. #3
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    Quote Originally Posted by takermaster35 View Post
    y^2=x^3(1-x)^6 . How to find the area of this equation. Thank you in advance
    Probably, you want to find the area between the curve and the x-axis.

    This is y=\pm\sqrt{x^3(1-x)^6}

    Above the x-axis it's y=\sqrt{x^3(1-x)^6}

    This is zero when x=0 and when x=1. For negative x, this would be complex.
    Beyond x=1, the curve travels to infinity.

    Therefore, integrate y=\sqrt{x^3(1-x)^6} from x=0 to x=1,
    and double your result if you want to include the region below the x-axis.

    You can start by simplifying the powers of x.
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