how to integrate x squared - x cube dx with a limit from 0 to 1?

\[\int\limits_{}^{}x^2-x^3dx\] yes?

Woops forgot to put in the limits of integration. My bad.

Anyways, remember that you can integrate two functions which are added/subtracted by first finding the integral of each, then adding/subtracting them (by the same token as differentiation of the sum/difference of two fxns) First, split this into two different integrals, then solve each. Piece them back together, and you'll have your F(x). After that, don't forget to plug in your limits of integration. Your answer should be \[F(x) = x^3/3 - x^4/4\] from 0 to 1 1/3 - 1/4 = 1/12

sorry, my fault.. its actually square root of x - x cube..

OK so it's \[\int\limits_{0}^{1} \sqrt{x-x^3}\]

only the first x has the square root :c

Oh ok that makes sense haha. I was scratching my head. Same deal though! Integrate each one separately then calculate each against the limits of integration.

wait, ill do it :D

\[\int\limits_{}^{} \sqrt{x} = (2x^{3/2})/3\] and \[\int\limits_{}^{}x^3 = x^4/4\] Combine the functions, take your limits of integration, and solve! Gotta get to bed though, good luck on this problem. If you still aren't getting it right, keep at it, there will be someone around to help ya ;)

thanks for the help dude :D, actually im just on the first step.. hehehe.. im doing to find the center of the mass.. hehehe

No problem! Good luck with the rest of the problem, but by now I'm sure it's been long since you solved it!

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