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Math Help - The anti-derivative law

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    The anti-derivative law

    Hi, I need to use the anti-derivative for the function of f(x) = -4/3(x^3-1). I need to get to the first function equation. However, there is a slight, and by slight I mean major, issue. I do not know and was not taught how to use the anti-derivative law with fractions. I keep on getting closer to the answer but cannot get it. This is what I got for f(x') = 2/3(x^-3 - 1)^2.

    What should I do next? f(x') is clearly incorrect.

    Edit. I also got f(x') = 1/2(x^3 1)^-4. This is much closer, but still wrong.
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    Re: The anti-derivative law

    Quote Originally Posted by Barthayn View Post
    Hi, I need to use the anti-derivative for the function of f(x) = -4/3(x^3-1). I need to get to the first function equation. However, there is a slight, and by slight I mean major, issue. I do not know and was not taught how to use the anti-derivative law with fractions. I keep on getting closer to the answer but cannot get it. This is what I got for f(x') = 2/3(x^-3 - 1)^2.

    What should I do next? f(x') is clearly incorrect.
    Assuming that your integral is \displaystyle \int{-\frac{4}{3(x^3 - 1)}\,dx} ...

    \displaystyle \begin{align*} \int{-\frac{4}{3(x^3 - 1)}\,dx} &= -\frac{4}{3}\int{\frac{1}{x^3 - 1}\,dx} \\ &= -\frac{4}{3}\int{\frac{1}{(x - 1)(x^2 + x + 1)}\,dx} \end{align*}

    You will now need to use Partial Fractions. See how you go.
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    Re: The anti-derivative law

    Quote Originally Posted by Prove It View Post
    Assuming that your integral is \displaystyle \int{-\frac{4}{3(x^3 - 1)}\,dx} ...

    \displaystyle \begin{align*} \int{-\frac{4}{3(x^3 - 1)}\,dx} &= -\frac{4}{3}\int{\frac{1}{x^3 - 1}\,dx} \\ &= -\frac{4}{3}\int{\frac{1}{(x - 1)(x^2 + x + 1)}\,dx} \end{align*}

    You will now need to use Partial Fractions. See how you go.
    My equation I am trying to solve is this f(x) = (-4/3) (x^3-1), where -4/3 is the constant of the function. Also. The teacher only taught the class how to use the law in the basic way. IE how to use it for the function of x^3 - 4x^2
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    Re: The anti-derivative law

    Quote Originally Posted by Barthayn View Post
    My equation I am trying to solve is this f(x) = (-4/3) (x^3-1), where -4/3 is the constant of the function. Also. The teacher only taught the class how to use the law in the basic way. IE how to use it for the function of x^3 - 4x^2
    OK, this is why it helps to learn a little basic LaTeX, so that everyone knows what you are trying to say...

    Anyway...

    \displaystyle \int{-\frac{4}{3}\left(x^3 - 1\right) \,dx} = -\frac{4}{3}\int{x^3 - 1\,dx}.

    Now integrate each of those terms, and multiply your answer by \displaystyle -\frac{4}{3}.
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    Re: The anti-derivative law

    I am still not getting the correct answer. I have no idea what to do because I have a variable with an exponent on the function. Anyways, the function should be f(x) = (-4/3) (x^(-3)-1). In other words, the function that has a variable on it should be negative and not positive.
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    Re: The anti-derivative law

    Quote Originally Posted by Barthayn View Post
    I am still not getting the correct answer. I have no idea what to do because I have a variable with an exponent on the function. Anyways, the function should be f(x) = (-4/3) (x^(-3)-1). In other words, the function that has a variable on it should be negative and not positive.
    I have now had to waste two posts telling you how to go about answering your question because I have tried to translate (wrongly) what you have written. I suggest you learn some basic LaTeX so that you can avoid frustrating the people here who are trying to help you.

    Now, assuming that your integral is \displaystyle \int{-\frac{4}{3}\left(x^{-3} - 1\right)\,dx}...

    \displaystyle \begin{align*} \int{-\frac{4}{3}\left(x^{-3} - 1\right)\,dx} &= -\frac{4}{3}\int{x^{-3} - 1\,dx} \end{align*}

    Now integrate that stuff term by term using the rule you know \displaystyle \int{x^n\,dx} = \frac{1}{n + 1}x^{n+1}. It doesn't make any difference if your exponent is negative (except if your exponent is -1).
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    Re: The anti-derivative law

    Quote Originally Posted by Prove It View Post
    I have now had to waste two posts telling you how to go about answering your question because I have tried to translate (wrongly) what you have written. I suggest you learn some basic LaTeX so that you can avoid frustrating the people here who are trying to help you.

    Now, assuming that your integral is \displaystyle \int{-\frac{4}{3}\left(x^{-3} - 1\right)\,dx}...

    \displaystyle \begin{align*} \int{-\frac{4}{3}\left(x^{-3} - 1\right)\,dx} &= -\frac{4}{3}\int{x^{-3} - 1\,dx} \end{align*}

    Now integrate that stuff term by term using the rule you know \displaystyle \int{x^n\,dx} = \frac{1}{n + 1}x^{n+1}. It doesn't make any difference if your exponent is negative (except if your exponent is -1).
    From \displaystyle \int{x^n\,dx} = \frac{1}{n + 1}x^{n+1}

    I get this:
    \displaystyle \int{f(x')\,dx} =  \frac{-4}{3}*\frac {x^{-2}}{-2}-1



    Now, I have to solve for x=2 and it should be \frac{5}{6}. But I get \frac{-5}{6}, which is wrong.

    I do not see where I am going wrong.
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    Re: The anti-derivative law

    Notice that:
    \frac{-4}{3} \int (x^{-3}-1)dx=\frac{-4}{3} \int x^{-3}dx - \left(\frac{-4}{3}\right)\int dx=...
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    Re: The anti-derivative law

    Thanks, for all your help, but I figured it out at school today. It turns out that you increase the exponent on all variables that has an exponent on it.
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    Re: The anti-derivative law

    Quote Originally Posted by Barthayn View Post
    It turns out that you increase the exponent on all variables that has an exponent on it.
    Yes, that's true altough if the exponent is -1 that's an exception.
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  11. #11
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    Re: The anti-derivative law

    \int \frac{-4}{3(x^3-1)}dx

    =\int \frac{1}{3(x-1)}dx - \int \frac{x+2}{3(x^2+x+1)}dx

    =\frac{1}{3}\ln{(x-1)} - \int \frac{x+2}{3(x^2+x+1)}dx
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