Evaluate the following integral: `int_1^2x\ sqrt(3x-2)\ dx`

Explanation

Given:`int_1^2 x sqrt(3x-2)\ dx`

`I=int_1^2 x sqrt(3x-2)\ dx`

Let `3x-2=t^2 implies dx=2tdt` and `3x=2+t^2 implies x=(t^2+2)/3`

When `x=1, t=1`and when `x=2, t=2`

`I=int_1^2 ((t^2+2)/3)txx2/3xxtdt`

`I=2/9 int_1^2 (t^4+2t^2)dt`

`I=2/9[(t^5)/5+2(t^3)/3]_1^2`

`I=326/135`