8.4k views
0 votes
Use the Comparison Theorem to determine whether the integral is convergent or divergent.

Use the Comparison Theorem to determine whether the integral is convergent or divergent-example-1
User FiniteRed
by
8.4k points

1 Answer

1 vote
First, examine what happens over the integration domain.


\displaystyle(\sin^2x)/(\sqrt x)=x^(3/2)(\sin^2x)/(x^2)\implies\lim_(x\to0^+)x^(3/2)(\sin^2x)/(x^2)=0*1=0


(\sin^2x)/(\sqrt x) is continuous for
x\\eq0, so we don't need to worry about the right endpoint.

Now, note that


(\sin^2x)/(\sqrt x)\le\frac1{\sqrt x}

due to the fact that
|\sin x|\le1\implies0\le\sin^2x\le1. So, if the larger integral converges, so must the smaller one. You have


\displaystyle\int_0^\pi(\mathrm dx)/(\sqrt x)=\lim_(t\to0^+)\int_t^\pi x^(-1/2)\,\mathrm dx

=\displaystyle\lim_(t\to0^+)2x^(1/2)\bigg|_(x=t)^(x=\pi)=2\left(\sqrt\pi-\lim_(t\to0^+)\sqrt t\right)=2\sqrt\pi

so the larger integral converges. Therefore the smaller one does, too.
User Cdsln
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.