# Integrating the root of a power

• Mar 29th 2011, 11:03 AM
djnorris2000
Integrating the root of a power
I'm working from a book and trying to integrate $\sqrt[4]{2x^{3}}$, unfortunately I always arrive at the wrong answer. Can anyone point out where I'm going wrong with this working?

$\sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}$

Therefore

$\frac{dy}{dx}=\frac{3}{4}(2x)^\frac{-1}{4}=\frac{3}{4}(\frac{1}{\sqrt[4]{2x}})$
• Mar 29th 2011, 11:11 AM
Plato
Quote:

Originally Posted by djnorris2000
I'm working from a book and trying to integrate $\sqrt[4]{2x^{3}}$,

$\displaystyle\int {x^{\frac{n}
{m}} dx} = \frac{m}
{{m + n}}x^{\frac{{m + n}}
{m}} + c$
• Mar 29th 2011, 11:50 AM
djnorris2000
Hmm, I've not come across that rule before, so would it give $\frac{4}{7}x^\frac{7}{4}+2$ or have I completely misinterpreted it?

I was trying to use $x^n=nx^{n-1}$.
• Mar 29th 2011, 11:54 AM
Plato
Quote:

Originally Posted by djnorris2000
I was trying to use $x^n=nx^{n-1}$.

Are you doing a derivative or an anti-derivative?
In the title of the OP it says integrate which is an anti-derivative.
• Mar 29th 2011, 01:53 PM
Quacky
Quote:

Originally Posted by djnorris2000
I'm working from a book and trying to integrate $\sqrt[4]{2x^{3}}$, unfortunately I always arrive at the wrong answer. Can anyone point out where I'm going wrong with this working?

$\sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}$

Therefore

$\frac{dy}{dx}=\frac{3}{4}(2x)^\frac{-1}{4}=\frac{3}{4}(\frac{1}{\sqrt[4]{2x}})$

Quote:

Originally Posted by djnorris2000
Hmm, I've not come across that rule before, so would it give $\frac{4}{7}x^\frac{7}{4}+2$ or have I completely misinterpreted it?

I was trying to use $x^n=nx^{n-1}$.

Quote:

Originally Posted by Plato
Are you doing a derivative or an anti-derivative?
In the title of the OP it says integrate which is an anti-derivative.

Regardless of the confusion, there is a gaping error in the OP, where you say that:

$\sqrt[4]{2x^{3}}=(2x)^\frac{3}{4}$

and I think you mean:

$\sqrt[4]{2x^{3}}=\sqrt[4]{2}\times(x)^\frac{3}{4}$ - the 2 isn't cubed. This should help with whatever it is you are trying to solve. I don't know whether you're trying to differentiate or integrate still, so please specify - copy the exact wording of the question.
• Mar 30th 2011, 09:54 AM
djnorris2000
Thanks Quacky, I meant differentiate all along. That's what I get for browsing ahead in the book and posting when I'm brain dead from work.

I figured I'd be making a simple mistake in there and now that you've pointed that out I'm arriving at the right answer.