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does the integral of 1/x+ srqtx from 1 to infinity converge?
 one year ago
 one year ago
does the integral of 1/x+ srqtx from 1 to infinity converge?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
the degree of the denominator has to be larger than the degree of the numerator by more than 1
 one year ago

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

electrokidBest 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!!
 one year ago

electrokidBest ResponseYou've already chosen the best response.0
so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
neither does the one above
 one year ago

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