Question

(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?

Answer 1

you could do a substitution
let u=1+x^3

Answer 2

\[\int\limits_{0}^{\infty}x^2/\sqrt{1+x^3}\]

Answer 3

without solving...
I think you can do comparison test

Answer 4

whats the comparison test?

Answer 5

http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegralsCompTest.aspx

Answer 6

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.

Answer 7

\[
\int_0^\infty\frac{x^2}{\sqrt{1+x^3}}\,dx\\
u=1+x^3\\
du=3x^2\,dx\\
=\frac{1}{3}\int_1^\infty\frac{1}{\sqrt{u}}\,du\\
=\frac{2}{3}\left[\sqrt{u}\right]_1^\infty\\
=\infty
\]

Answer 8

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.

Answer 9

but if you wanted me to find him a function... this is the one I would have chose..
\[\sqrt{1+x^3}<\sqrt{x^4} \\ \frac{1}{\sqrt{1+x^3}} > \frac{1}{\sqrt{x^4}} \\ \frac{x^2}{\sqrt{1+x^3}}> \frac{x^2}{\sqrt{x^4}}=\frac{x^2}{x^2}=1 \\ \text{ since }
\int\limits_0^\infty 1 dx \text{ diverges then } \int\limits_0^\infty \frac{x^2}{\sqrt{1+x^3}} dx \text{ diverges }\]

Answer 10

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 \(x\geq2\).

Answer 11

I should have probably included x^4>1+x^3 for x>=2 above

Answer 12

we know
\[\int\limits_0^\infty \frac{x^2}{\sqrt{1+x^3}} dx=\int\limits_0^2 \frac{x^2}{\sqrt{1+x^3}} dx+\int\limits_2^\infty \frac{x^2}{\sqrt{1+x^3}} \\ \text{ you will have a number }+\text{ something that diverges } \\ \text{ which still mean the inetgral diverges }\]

Answer 13

2 is certainly not the smallest value when \(x^4>1+x^3\) but that will do.

Answer 14

thanks for help and the link as well.