Integral Convergence

[halcyon schooner]

I have this task of testing if the integral converges.

I'm not sure why its ' > ' , shouldn't it be '<'.
Their conclusion is that it diverges since 1/x diverges.

[copper bone]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[cursive vine]

$\abs{1+2x^3\cos^2(x)}\leq1+2x^3\cos^2(x)\leq1+2x^3$ by triangle inequality, then the fact $\cos^2(x)\leq 1$

[split imp]

Omegabet_

[cursive vine]

taking reciprocal will then switch the inequality

[halcyon schooner]

my idea was if (1+x^3) > x^3
then the reciprocal is just '<'

and thats completely different from theirs

[cursive vine]

making denominators smaller increases the overall value

oh you mean the last inequality

yeah the last one is wrong

[halcyon schooner]

0 <= cos(x)^2 <= 1 right

so it doesnt matter if it's in the abs

[cursive vine]

yes but you're not talking about that

you're talking about the 2nd inequality

apparently, you never actually said which for some reason

[halcyon schooner]

the whole inequalitiy where its '>' didn't make sense to me

[cursive vine]

they're both >

as I said

the last one is wrong

[halcyon schooner]

by this isnt the x^2/ (1+x^3) < x^2/x^3

[cursive vine]

yes, like I said already....

the last one is wrong.......

[cursive vine]

.

[halcyon schooner]

got it

[cursive vine]

the 1st one is correct from graphing them.

[halcyon schooner]

could this be done if we use the second criterion of convergence of the non-property integral, and find the limit of f(x) / 1/x which would prove that this integral is divergent since integral 1/x is divergent on this range 1 to infinity

[halcyon schooner]

+close