# Integration help

• Jan 8th 2008, 02:49 PM
nugiboy
Integration help
Hi i need to integrate this function:

http://i137.photobucket.com/albums/q...iger/Integ.jpg

I can do the first bit as $x^0.5 goes to 2/3x^3/2$

Im really stuck on the second $6/x^3$. I need to somehow convert that into a form where i can integrate it. ie, not a fraction. Any suggestions?
• Jan 8th 2008, 02:51 PM
Jhevon
Quote:

Originally Posted by nugiboy
Hi i need to integrate this function:

http://i137.photobucket.com/albums/q...iger/Integ.jpg

I can do the first bit as $x^0.5 goes to 2/3x^3/2$

Im really stuck on the second $6/x^3$. I need to somehow convert that into a form where i can integrate it. ie, not a fraction. Any suggestions?

write it as $6x^{-3}$

now what?

and if your power has more than one character, you must keep them in {} brackets, that is, type x^{0.5} and x^{3/2} as opposed to what you typed. also, you should be aware that LaTeX removes spaces, so that's why what you typed looked so messy. use ~ to insert spaces, or type your text messages in a \text {} or \mbox {}
• Jan 8th 2008, 02:58 PM
Peritus
(2/3)*x^(3/2) + (-1/2)*6*(x^(-2)) = (2/3)*x^(3/2) -3*(x^(-2))
• Jan 8th 2008, 02:59 PM
nugiboy
Quote:

Originally Posted by Jhevon
write it as $6x^{-3}$

now what?

and if your power has more than one character, you must keep them in {} brackets, that is, type x^{0.5} and x^{3/2} as opposed to what you typed. also, you should be aware that LaTeX removes spaces, so that's why what you typed looked so messy. use ~ to insert spaces, or type your text messages in a \text {} or \mbox {}

Hi thanks for the reply and the help on the Latex.

Wouldn't $6x^{-3}$ be $1/{6x^3}$? Which is different to $6/{x^3}$
• Jan 8th 2008, 03:03 PM
Jhevon
Quote:

Originally Posted by nugiboy
Hi thanks for the reply and the help on the Latex.

Wouldn't $6x^{-3}$ be $1/{6x^3}$? Which is different to $6/{x^3}$

no. that would be $\frac 16x^{-3} = \frac 1{6x^3}$, the -3 power only applies to the x, the 6 has nothing to do with it.
• Jan 8th 2008, 03:06 PM
nugiboy
Quote:

Originally Posted by Jhevon
no. that would be $\frac 16x^{-3} = \frac 1{6x^3}$, the -3 power only applies to the x, the 6 has nothing to do with it.

Ohhhh right. Thanks it all makes sense now!