anonymous
  • anonymous
(Improper Integrals) Determine wither the integral is convergent or divergent. Evaluate if it's convergent. infinity ∫x^2/√(1+x^3) 0 how would you know straight away that this integral is divergent without actually solving for it?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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freckles
  • freckles
you could do a substitution let u=1+x^3
anonymous
  • anonymous
\[\int\limits_{0}^{\infty}x^2/\sqrt{1+x^3}\]
freckles
  • freckles
without solving... I think you can do comparison test

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anonymous
  • anonymous
whats the comparison test?
freckles
  • freckles
http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx
thomas5267
  • thomas5267
By comparing with what? My guess is that this integral does not converge since \[ \frac{x^2}{\sqrt{1+x^3}}\approx\frac{x^2}{x^{3/2}}=x^{1/2}\\ \int_0^\infty x^{1/2}\,dx=\infty \] Note that this is not rigorous at all.
thomas5267
  • thomas5267
\[ \int_0^\infty\frac{x^2}{\sqrt{1+x^3}}\,dx\\ u=1+x^3\\ du=3x^2\,dx\\ =\frac{1}{3}\int_1^\infty\frac{1}{\sqrt{u}}\,du\\ =\frac{2}{3}\left[\sqrt{u}\right]_1^\infty\\ =\infty \]
freckles
  • freckles
Am I being asked "By comparing with what?" ? If he wants to use the comparison test, then he will have to find a function to compare his function with... The comparison test is stated on that one link I provided.
freckles
  • freckles
but if you wanted me to find him a function... this is the one I would have chose.. \[\sqrt{1+x^3}<\sqrt{x^4} \\ \frac{1}{\sqrt{1+x^3}} > \frac{1}{\sqrt{x^4}} \\ \frac{x^2}{\sqrt{1+x^3}}> \frac{x^2}{\sqrt{x^4}}=\frac{x^2}{x^2}=1 \\ \text{ since } \int\limits_0^\infty 1 dx \text{ diverges then } \int\limits_0^\infty \frac{x^2}{\sqrt{1+x^3}} dx \text{ diverges }\]
thomas5267
  • thomas5267
Yes I am asking you compare the function with what. Sorry for the rudeness. It simply didn't occur to me that \(x^4>1+x^3\) for \(x\geq2\).
freckles
  • freckles
I should have probably included x^4>1+x^3 for x>=2 above
freckles
  • freckles
we know \[\int\limits_0^\infty \frac{x^2}{\sqrt{1+x^3}} dx=\int\limits_0^2 \frac{x^2}{\sqrt{1+x^3}} dx+\int\limits_2^\infty \frac{x^2}{\sqrt{1+x^3}} \\ \text{ you will have a number }+\text{ something that diverges } \\ \text{ which still mean the inetgral diverges }\]
thomas5267
  • thomas5267
2 is certainly not the smallest value when \(x^4>1+x^3\) but that will do.
anonymous
  • anonymous
thanks for help and the link as well.

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