Question

Does this diverge or converge?

Answer 1

\[\LARGE \int_{0}^{\infty}~\frac{x^2}{\sqrt{1+x^3}}~dx\]

Answer 2

integrate and check

Answer 3

Good question. So the best way to see this is to actually integrate

Answer 4

Would I set it up as:
\[\LARGE lim_{x \rightarrow \infty}\int_{0}^{t}~\frac{x^2}{\sqrt{1+x^3}}~dx\] ?

Answer 5

I just learned this today btw. Like a few hours ago xD

Answer 6

nah just integrate it like normal and sub in the bounds use U sub u=1+x^3

Answer 7

So it should be achievable with the u-sub u=x^3
Better yet if you can do it by inspection

Answer 8

@Luigi0210 almost yes
limit as t->infinity --That's pretty much how you handle all these improper integrals. You replace the infinity with a parameter and take the limit as the parameter goes to infinity

Answer 9

This is just a bit of technical machinery to avoid the "non-sensicalness" of evaluating a function at "infinity"

Answer 10

It makes more sense if you start to write down epsilon-delta type formulations for the integrals

Answer 11

Oh, I didn't see that mistake >_>

but couldn't we find out using \(\frac{1}{x}\) and \(\frac{1}{x^2}\)