Question

Integation (converges or diverges)

Answer 1

what is the easiest way to show this integral converges or diverges?
\[\int\limits_{0}^{1}\frac{ 1 }{ \sqrt{x+x^3} }dx\]
I know the limit comparison test and the comparison test.

I also know
\[\int\limits\limits_{0}^{1}\frac{ 1 }{ x^p }dx\]
converges when p<1 and diverges when p≥1

Answer 2

@reemii do it again, friend. I know you can do it.

Answer 3

near zero, the bigger term is \(x\), not \(x^3\), so my guess would be that the integral converges.

how to do that. hm.
\(\int_\epsilon^1 \frac{1}{\sqrt{x+x^3}}dx \le \int_\epsilon^1 \frac1{\sqrt x}dx\) for all epsilon.
and the limit , as epsilon -> 0 , of the second integral is a finite value.

Answer 4

I cant see why I can compare with\[\frac{ 1 }{ \sqrt{x} }\]
if you see on the attachment picture\[\int\limits_\epsilon^1 \frac{1}{\sqrt{x+x^3}}dx \le \int\limits_\epsilon^1 \frac1{\sqrt x}dx\] this is not true for all 0≤x≤1

Answer 5

the epsilon is taken between 0 and 1. (0 excluded)

Answer 6

the convergence of the integral equivalent to the convergence of the limit.
since we prove by this inequality that the bigger integral converges, your integral converges.

Answer 7

Sorry.. Of course..