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

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Answer 1 · 2018-05-21T15:09:30

The answer is #=6#

The improper integral is

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

There is an undefined point when #x=0#

Therefore,

#int_-1^1x^(-2/3dx)=lim_(p->0)int_-1^px^(-2/3)dx+lim_(p->0)int_p^1x^(-2/3)dx#

#lim_(p->0)int_-1^px^(-2/3)dx=lim_(p->0)[3x^(1/3)]_-1^p#

#=lim_(p->0)(3p^(1/3)+3)#

#=3#

#lim_(p->0)int_p^1x^(-2/3)dx=lim_(p->0)[3x^(1/3)]_p^1#

#=lim_(p->0)(3-3p^(1/3))#

#=3#

Finally,

#int_-1^1x^(-2/3dx)=3+3=6#