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Thread: Integration Question

  1. #1
    Member looi76's Avatar
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    Integration Question

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

    Attempt:

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

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

    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 $\displaystyle u=6-2x$, for example.
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  3. #3
    GAMMA Mathematics
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    Quote Originally Posted by looi76 View Post
    Question
    A curve is such that $\displaystyle \frac{dy}{dx} = \frac{4}{\sqrt(6 - 2x)}$, and $\displaystyle P(1,8)$ is a point on the curve. Find the equation of the curve.

    Attempt:

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

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

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

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

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

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

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

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

    The important part is that you knew what to do!
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