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

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Answer 1 · 2017-02-22T22:46:08

#int_3^oo (x^-2-x^-3)dx = 5/18#

We have:

#int_3^oo (x^-2-x^-3)dx = lim_(u->oo) int_3^u (x^-2-x^-3)dx#

Using the power rule:

#int_3^u (x^-2-x^-3)dx = [-x^-1 +x^-2/2]_3^u = 1/(2u^2)-1/u +1/3-1/18 = 1/u(1/(2u)-1) +5/18#

So:

#int_3^oo (x^-2-x^-3)dx = lim_(u->oo) 1/u(1/(2u)-1) +5/18 = 0*(-1)+5/18 = 5/18#