Let XXX be continuous with PDF f(x)=kx−3f(x) = k x^{-3}f(x)=kx−3 for x≥1x \ge 1x≥1. Find kkk and then compute E[X]E[X]E[X].
k=2,E[X]=2k=2, E[X]=2k=2,E[X]=2
k=2,E[X]=∞k=2, E[X]=\inftyk=2,E[X]=∞
k=1,E[X]=2k=1, E[X]=2k=1,E[X]=2
k=12,E[X]=∞k=\frac{1}{2}, E[X]=\inftyk=21,E[X]=∞