How do you evaluate the definite integral #int (3sqrtx) dx# from #[1,3]#?

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Answer 1 · 2016-12-12T15:31:10

The answer is #=2(3sqrt3-1)=8.39#

We use

#intx^ndx=x^(n+1)/(n+1)+C# #(n!=-1)#

So,

#int_1^3 3sqrtxdx=3int_1^3x^(1/2)dx#

#=3 [x^(3/2) /(3/2) ] _1^3 #

#=2 [3^(3/2)-1] #

#=2(3sqrt3-1)#

#=8.39#

Answer 2 · 2016-12-12T16:03:16

#int_1^3 3sqrt(x)dx = 2 (sqrt(27) -1)#

#int_1^3 3sqrt(x)dx = 3 int_1^3 x^(1/2)dx =3 [ x^(3/2)/(3/2)]_1^3 = 2 [ x^(3/2)]_1^3 = 2 (sqrt(27) -1)#