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Math Help - Fundamental Theorm

  1. #1
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    Fundamental Theorm

    Find the derivative of the function.


    I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

    What I am talking about is breaking the original equation down to...

    \int_5x^0 frac_((u^2)-1)/((u^2)+1) +\int_0^6x frac_((u^2)-1)/((u^2)+1)

    Or is that completely wrong?

    Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry
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  2. #2
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    ryan18:

    Yes that's fine; that's called the Additive Property of Integrals. Are you deriving or integrating? To integrate that function use partial fractions.
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  3. #3
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    Quote Originally Posted by ryan18 View Post
    Find the derivative of the function.


    I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

    What I am talking about is breaking the original equation down to...

    \int_5x^0 frac_((u^2)-1)/((u^2)+1) +\int_0^6x frac_((u^2)-1)/((u^2)+1)

    Or is that completely wrong?

    Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry

    \frac{u^2-1}{u^2+1}=1-\frac{2}{u^2+1}\Longrightarrow \int\limits^{6x}_{5x}\frac{u^2-1}{u^2+1}\,dx=\left[u-2\arctan u\right]^{6x}_{5x} =6x-2\arctan 6x-5x+2\arctan 5x , and now make some algebraic order here and differentiate.

    Tonio
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  4. #4
    Junior Member eddie2042's Avatar
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    Quote Originally Posted by ryan18 View Post
    Find the derivative of the function.


    I was watching a video on how to seperate them and they used 0. Can I do that with these two functions?

    What I am talking about is breaking the original equation down to...

    \int_5x^0 frac_((u^2)-1)/((u^2)+1) +\int_0^6x frac_((u^2)-1)/((u^2)+1)

    Or is that completely wrong?

    Wow so I suck at making integrals...but the first one is 5x on the bottom and 0 on top. the second is 0 on bottom and 6x on top....sorry

    Yes you can use any constant at all; won't make any difference. But I will just use a constant C.

    g(x) = \int_{5x}^{c}\frac{u^2 - 1}{u^2 + 1}du + \int_{c}^{6x}\frac{u^2 - 1}{u^2 + 1}du

    but since the variable has to be as the upper limit, we'll put a minus sign in front of the first integral and switch the limits

    g(x) = -\int_{c}^{5x}\frac{u^2 - 1}{u^2 + 1}du + \int_{c}^{6x}\frac{u^2 - 1}{u^2 + 1}du

    Then the derivative will be

    \frac{dg(x)}{dx} = -5\frac{25x^2 - 1}{25x^2 + 1} + 6\frac{36x^2 - 1}{36x^2 + 1}

    Also, practice your LaTeX typing skills.. it'll come in handy =P
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