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Math Help - Integration Question

  1. #1
    Member looi76's Avatar
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    Integration Question

    Question
    A curve is such that \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}, and P(1,8) is a point on the curve. Find the equation of the curve.

    Attempt:

    \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}

    \int 4(6-2x)^{-\frac{1}{2}}dx

    I don't know how to integrate it.. need help!
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  2. #2
    Super Member wingless's Avatar
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    Substitute u=6-2x, for example.
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  3. #3
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by looi76 View Post
    Question
    A curve is such that \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}, and P(1,8) is a point on the curve. Find the equation of the curve.

    Attempt:

    \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}

    \int 4(6-2x)^{-\frac{1}{2}}dx

    I don't know how to integrate it.. need help!
    Let u = 6 - 2x, du = -2 dx, dx = -0.5 du

    \int 4(6-2x)^{-\frac{1}{2}}dx

    \int -2(u)^{-\frac{1}{2}}du

    -4(u)^{\frac{1}{2}}

    y = -4(6 - 2x)^{\frac{1}{2}} + C

    Solve for C by knowing that y(1)=8

    The important part is that you knew what to do!
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