Find` ∫ 7 log⁡x.x dx`
`a) 7/2(logx−x)+C`
`b) –7/2(x^2logx−x^3)+C`
`c) 7/2(x^2logx−x)+C`
`d) (x^2 log⁡x+x)+C`

Explanation

Consider the line perpendicular to the plane of the ure through the centre of the cross. The moment of inertia of each rod about this line is `(Ml^2)/12` and hence the moment of inertia of the cross is `(Ml^2)/6`. The moment of inertia of the cross about the two bisectors are equal by symmetry and according to the theorem of perpendicular axes the moment of inertia of the cross about the bisector is `(Ml^2)/12`