How do you evaluate the definite integral #int x^2sqrt(1-x^3)# from #[0,1]#?

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Answer 1 · 2016-09-07T18:00:48

#= 2/9#

spot the pattern

#d/dx (f^(x))^n = n (f(x))^(n-1) f'(x)#

So

#int_0^1 x^2sqrt(1-x^3) dx#

#= int_0^1 d/dx(-2/9 root3(1-x^3) )dx#

#= -2/9[ root3(1-x^3) ]_0^1#

#= -2/9 (0 - 1) = 2/9#