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peggiepenguin
 2 years ago
does the integral of 1/x+ srqtx from 1 to infinity converge?
peggiepenguin
 2 years ago
does the integral of 1/x+ srqtx from 1 to infinity converge?

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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1the degree of the denominator has to be larger than the degree of the numerator by more than 1

electrokid
 2 years ago
Best ResponseYou've already chosen the best response.0is it \[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\] OR \[\int_1^\infty{dx\over x+\sqrt{x}} \]

electrokid
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73 aha... @peggiepenguin next time please use the "equation editor on this website" or "TEX" to enter your equations.. so we know what you are talking about!!

electrokid
 2 years ago
Best ResponseYou've already chosen the best response.0so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1neither does the one above

electrokid
 2 years ago
Best ResponseYou've already chosen the best response.0aah.. jumping to conclusions too soon. \[ u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\ \int_1^\infty\frac{2u}{u^2+u}du\\ =2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty} \]
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