# The anti-derivative law

• Sep 22nd 2011, 09:25 PM
Barthayn
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. (Headbang)
• Sep 22nd 2011, 09:28 PM
Prove It
Re: The anti-derivative law
Quote:

Originally Posted by Barthayn
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.
• Sep 22nd 2011, 09:36 PM
Barthayn
Re: The anti-derivative law
Quote:

Originally Posted by Prove It
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
• Sep 22nd 2011, 09:54 PM
Prove It
Re: The anti-derivative law
Quote:

Originally Posted by Barthayn
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 $\displaystyle -\frac{4}{3}$.
• Sep 22nd 2011, 10:10 PM
Barthayn
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.
• Sep 22nd 2011, 10:30 PM
Prove It
Re: The anti-derivative law
Quote:

Originally Posted by Barthayn
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).
• Sep 23rd 2011, 05:03 AM
Barthayn
Re: The anti-derivative law
Quote:

Originally Posted by Prove It
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.
• Sep 23rd 2011, 05:30 AM
Siron
Re: The anti-derivative law
Notice that:
$\frac{-4}{3} \int (x^{-3}-1)dx=\frac{-4}{3} \int x^{-3}dx - \left(\frac{-4}{3}\right)\int dx=...$
• Sep 23rd 2011, 02:32 PM
Barthayn
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.
• Sep 24th 2011, 12:30 AM
Siron
Re: The anti-derivative law
Quote:

Originally Posted by Barthayn
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.
• Sep 24th 2011, 07:33 AM
sbhatnagar
Re: The anti-derivative law
$\int \frac{-4}{3(x^3-1)}dx$

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

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