strubenuff

I'm sorry to keep using these forums for help, but I can't find any active calclus forums online. If anyone knows of any, I'd greatly apprecitate them tell me. My problem is that I'm integrating 5x*(1-x²)^(1/3). I let u= (1-x²), so du= -2xdx. I think I just screwed up when I tried to compensate for the du...Anyhow, I had (-5/2)*the indenite integral of u^(1/3)du. Integrated, it's (-5/2)*(3u^(4/3))/4). Distributing gives us (-15u^(4/3))/8, and back substituting yields [-15(1-x²)^(4/3)]/8. Do you see a mistake?

enigma555

I got the same result. Looks good.

<a href="http://www.physicsforums.com" target="_blank">www.physicsforums.com</a> has a math section as well as a homework help section.

[ October 29, 2002: Message edited by: enigma555 ]</p>

godlessmath

Quick check:

Differentiate what you got as your answer and see if it is the same guy that you were integrating. If it is, then your answer is ok, if not, then something is up.