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let the rod is placed along x axis and its end A is at origin........
consider a small element of thickness dx at a distance x from end A..
mass of element is (mass per unit length)(length of this element)
dm=mxdx........1
moment of inertia of this element about the axis passing through its end A and perpendicular to rod is dmx^2
dI=dmx^2
=mx^3dx
total moment of inertia is integral of dI limit from 0 to 2L
I=integral(mx^3dx) lim 0 to 2L
=m (2L)^4 /4
I =4mL^2
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