Integrate` x^2/e^(x^3).`
`a) –1/(3e^(x^3))+C`
`b) 1/3e^(x^3)+C`
`c) –1/e^(x^3)+C`
`d) e^(x^3)+C`

Explanation

The situation is shown in ure. Let the speed of the masses be v at time t. This will also be the speed of a point on the rim of the wheel and hence the angular velocity of the wheel at time t will be v/r. If the height descended by the mass M is h the loss in the potential energy of the mass plus the pulley system is `mgh-mgh.` The gain in kinetic energy is `1/2Mv^2+1/2mv^2+1/2I(v/r)^2`. An no energy is lost,
`1/2(M=m+1/r^2)v^2=(M-m)gh`
`or, v^2=(2(M-m)gh)/(M+m+1/r^2)`
The angular momentum of the mass M is Mvr and that of the mass is mvr in the same direction. The angular momentum of the pulley is `Iomega=Iv/r.` The total angular momentum is
`[(M+m)r+I/r]v=[(M+m+I/r)r]sqrt((2(M-m)gh)/(M+m+I/r^2)) `
`=sqrt(2(M-m)(M+M+1/r^2)r^2gh)`