A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
(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?
anonymous
 one year ago
(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?

This Question is Closed

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you could do a substitution let u=1+x^3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{\infty}x^2/\sqrt{1+x^3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4without solving... I think you can do comparison test

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0whats the comparison test?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1By 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
 one year ago
Best ResponseYou've already chosen the best response.1\[ \int_0^\infty\frac{x^2}{\sqrt{1+x^3}}\,dx\\ u=1+x^3\\ du=3x^2\,dx\\ =\frac{1}{3}\int_1^\infty\frac{1}{\sqrt{u}}\,du\\ =\frac{2}{3}\left[\sqrt{u}\right]_1^\infty\\ =\infty \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4Am 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
 one year ago
Best ResponseYou've already chosen the best response.4but 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
 one year ago
Best ResponseYou've already chosen the best response.1Yes 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
 one year ago
Best ResponseYou've already chosen the best response.4I should have probably included x^4>1+x^3 for x>=2 above

freckles
 one year ago
Best ResponseYou've already chosen the best response.4we 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
 one year ago
Best ResponseYou've already chosen the best response.12 is certainly not the smallest value when \(x^4>1+x^3\) but that will do.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks for help and the link as well.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.