evaluate ⌠1 (x-x^2)/(2*x^(1/3))
⌡-8
(the signs with 1 and -8 is supposed to be an integral sign... w/ upper and lower bounds 1 & -8 respectively)

Question

evaluate ⌠1 (x-x^2)/(2*x^(1/3)) 
              ⌡-8
(the signs with 1 and -8 is supposed to be an integral sign... w/ upper and lower bounds 1 & -8 respectively)

anonymous · Answer

Divide the x^(1/3) through the numerator. Then integrate term by term.

anonymous · Answer

$$\int_{-8}^1\frac{x-x^2}{2x^{\frac{1}{3}}}dx$$

anonymous · Answer

Pull out the 1/2 then divide all the terms by x^(1/3) giving:
$$\frac{1}{2} \int\limits x^{2/3}-x^{5/3}dx=\frac{1}{2}(\frac{3}{5}x^{5/3}-\frac{3}{8}x^{8/3})+C$$

anonymous · Answer

Then evaluate it^^

anonymous · Answer

thanks! but when you plug in -8, i get an undefined number... is it okay to just take that out?

anonymous · Answer

when you evaluate it, would the answer be -383/10 ?

anonymous · Answer

H/o I'll tell you

anonymous · Answer

You actually get a complex answer D:

anonymous · Answer

http://www.wolframalpha.com/input/?i=integral+of+%28x-x^2%29%2F%282x^%281%2F3%29%29%2Cx%2C-8%2C1

anonymous · Answer

The problem is, you get a discontinuity at x=0. So you have to integrate it from -8 to y then y to 1. Then take the limit as y=>0

anonymous · Answer

thanks!

anonymous · Answer

No problem. Sorry I can't help you more than that :/