How do you find the definite integral for: #root3(x^2)# for the intervals [-1,1]?

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Answer 1 · 2018-04-11T14:42:21

#int_(-1)^1 root(3)(x^2)dx =6/5#

We know that:

#root(3)(x^2) = x^(2/3)#

and that in general:

#int x^adx = x^(a+1)/(a+1) + C#

#int_(-1)^1 root(3)(x^2)dx = int_(-1)^1 x^(2/3)dx#

#int_(-1)^1 root(3)(x^2)dx =[x^(2/3+1)/(2/3+1)]_(-1)^1#

#int_(-1)^1 root(3)(x^2)dx =[3/5 x^(5/3)]_(-1)^1#

#int_(-1)^1 root(3)(x^2)dx =3/5- (-3/5)#

#int_(-1)^1 root(3)(x^2)dx =6/5#