Re: The anti-derivative law

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**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 \displaystyle \int{-\frac{4}{3(x^3 - 1)}\,dx} $...

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

You will now need to use Partial Fractions. See how you go.

Re: The anti-derivative law

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**Prove It** Assuming that your integral is $\displaystyle \displaystyle \int{-\frac{4}{3(x^3 - 1)}\,dx} $...

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

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

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...

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

Now integrate each of those terms, and multiply your answer by $\displaystyle \displaystyle -\frac{4}{3}$.

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.

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

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

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

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

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

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

From $\displaystyle \displaystyle \int{x^n\,dx} = \frac{1}{n + 1}x^{n+1}$

I get this:

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

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

I do not see where I am going wrong.

Re: The anti-derivative law

Notice that:

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

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.

Re: The anti-derivative law

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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.

Re: The anti-derivative law

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

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

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