How do you test the improper integral #int (2x-1)^(-2/3)dx# from #[0,1]# and evaluate if possible?

Question

1291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-16T15:14:01

The integral converges to #=3#

The critical point is when #x=1/2#

Therefore,

The integral is

#int_0^1(2x-1)^(-2/3)dx #

#= lim_(t->(1/2))int_0^t(dx)/(2x-1)^(2/3)+lim_(t->(1/2))int_t^1(dx)/(2x-1)^(2/3)#

#= lim_(t->(1/2))[3/2(2x-1)^(1/3)]_0^t+ lim_(t->(1/2))[3/2(2x-1)^(1/3)]_t^1#

#= lim_(t->(1/2))(3/2*(2t-1)^(1/3)-3/2*(-1)^(1/3)) + lim_(t->(1/2))(3/2*(1)^(1/3)-3/2*(2t-1)^(1/3)) #

#=3/2+3/2#

#=3#