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

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

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

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