Integrate `xe^(2x)`.
`a) (e^(2x))/4(x−14)+C`
`b) (e^(2x))/4(2x−1)+C`
`c) (e^(2x))/2(2x−1)+C`
`d)( e^(2x))/4(x+1)+C`

Explanation

The torque acting on the mass m about the vertical axis through the hole is zero. The angular momentum about this axis, therefore, remains constant. If the speed of the mass is v whenit moves in the circel of radius r we have
`mv_0r_0=mvr`
or `v=r_0/rv_0`.......i ltbr. a. the tension `T=(mv^2)/r=(mr_0^2v_0^2)/r^3
b. The change in kinetic energy `=1/2mv^2-1/2mv_0^2`
Byi it is `=1/2mv_0^2[r_0^2/r^2-1]`