How do you find the definite integral of #int x^3-xdx# from #[0,2]#?

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Answer 1 · 2016-03-08T18:37:12

#int_0^2(x^3-x)dx=2#

#int_0^2(x^3-x)dx=[x^4/4-x^2/2]_0^2#

#=2^4/4-2^2/2-0#

#=2#

Answer 2 · 2016-03-08T18:38:55

2

use power integration formula:

#int x^ndx = x^(n+1) / (n+1) + c#

Hence:

#int (x^3-x) dx = x^4 / 4 - x^2 / 2 |_0^2 = 2^4 / 4 - 2^2 / 2 =2#