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Math Help - Comparing dy/dx with another dy/dx to find Y

  1. #1
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    Comparing dy/dx with another dy/dx to find Y

    Given that y=(3x+2)\sqrt{x+2} and \frac{dy}{dx}=\frac{3x+8}{2\sqrt{x+2}}.

    Integrate \int^7_{2}\frac{x}{2\sqrt{x+2}} with respect to x
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  2. #2
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    \frac{x}{2\sqrt{x+2}}= \frac{1}{3}\frac{3x+ 8}{2\sqrt{x+ 2}}- \frac{4}{3\sqrt{x+2}}
    You are "given" how to integrate the first and the second can be integrated by letting u= x+ 2.
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  3. #3
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    Quote Originally Posted by Punch View Post
    Given that y=(3x+2)\sqrt{x+2} and \frac{dy}{dx}=\frac{3x+8}{2\sqrt{x+2}}.

    Integrate \int^7_{2}\frac{x}{2\sqrt{x+2}} with respect to x
    Something's not right here. It's either your y or your \frac{dy}{dx} b/c

    \dfrac{d}{dx} (3x+2)\sqrt{x+2} = \dfrac{9x+14}{2\sqrt{x+2}}.
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    dy/dx=\frac{(3x+2)}{2\sqrt{x+2}}+3\sqrt{x+2}
    =\frac{9x+14}{2\sqrt{x+2}}

    you are absolutely right but i still dont get how i can compare \frac{9x+14}{2\sqrt{x+2}} with \frac{x}{2\sqrt{x+2}}
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  5. #5
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    First off, as HallsofIvy point out, the substitution u = x+2 or x = u - 2 will give you two integrals that you can easily integrate (and the way I personally would have done this problem). In what follows, the constants of integration will be supressed. First, using the fact that

    \displaystyle \int \dfrac{9x+14}{2\sqrt{x+2}}dx = (3x+2)\sqrt{x+2}

    then

    \displaystyle 9 \int \dfrac{x}{\sqrt{x+2}}dx + 14 \int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;(1)

    noting that what you want is the first integral. Also 9x+14 = 9(x+2) - 4 so your integral can be written as

    \displaystyle \int \dfrac{9(x+2)-4}{\sqrt{x+2}}dx = 9 \int \sqrt{x+2}\,dx  - 4\int \dfrac{1}{\sqrt{x+2}}dx

    so

    \displaystyle  9 \int \sqrt{x+2}\,dx  - 4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2},\;\;\;(2)


    Now use integration by parts \int u dv = uv - \int v du on the first integral in (2)

    u = \sqrt{x+2}, \;\; du = \dfrac{dx}{2\sqrt{x+2}}

    dv = dx,\;\;\; v = x

    gives

    \displaystyle 9 x \sqrt{x+2} - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx -  4\int \dfrac{1}{\sqrt{x+2}}dx = (6x+4)\sqrt{x+2}

    or

    \displaystyle  - \dfrac{9}{2}\int \dfrac{x}{\sqrt{x+2}}dx -  4\int \dfrac{1}{\sqrt{x+2}}dx = (-3x+4)\sqrt{x+2}

    or

    \displaystyle  9\int \dfrac{x}{\sqrt{x+2}}dx + 8\int \dfrac{1}{\sqrt{x+2}}dx = (6x-8)\sqrt{x+2}.\;\;(3)

    Now multiply 7 \times (3)  - 4 \times (1) gives

    \displaystyle  27 \int \dfrac{x}{\sqrt{x+2}}dx = 18(x-4)\sqrt{x+2}

    from which it follows that

    \displaystyle  \int \dfrac{x}{\sqrt{x+2}}dx = \frac{2}{3} (x-4)\sqrt{x+2}.

    Kinda long and again, not the way I would have done this problem.
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