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

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Answer 1 · 2017-12-04T11:13:55

The integral is divergent

The improper integral is calculated in #2# parts :

# lim_(q->0^+)int_q^1x^(-3/2)dx+lim_(p->oo)int_1^p x^(-3/2)dx#

First part

#lim_(q->0^+)int_q^1x^(-3/2)dx=lim_(q->0^+)[x^(-3/2+1)/(-3/2+1)]_q^1#

#=lim_(q->0^+)[x^(-1/2)/(-1/2)]_q^1#

#=lim_(q->0^+)[2x^(-1/2)]_1^q#

#=lim_(q->0^+)(2q^(-1/2)-2)#

#=lim_(q->0^+)(2/sqrtq-2)#

#=+oo#

This part is divergent, therefore the integral is divergent

Second part

#lim_(p->oo)int_1^p x^(-3/2)dx=lim_(p->oo)[x^(-3/2+1)/(-3/2+1)]_1^p#

#=lim_(p->oo)[x^(-1/2)/(-1/2)]_1^p#

#=lim_(p->oo)[2x^(-1/2)]_p^1#

#=lim_(p->oo)(2-2p^(-1/2))#

#=lim_(p->oo)(2-2/sqrtp)#

#=2#

This part is convergent.