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Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\int_0^1\frac{dx}{\sqrt{2xx^2}}\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
improper intergral
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Not sure how much that helps but: \[\Large x^2+2x=(x^22x1+1)=(x1)^2+1 \]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[\Large \int\frac{dx}{\sqrt{1(x1)^2}} \]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i think you are not going to find a nice closed form for this one, although you can if you try very hard if you are not trying to actually compute the integral, but only say whether or not it converges, then it is much easier
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
yes we dont really want to compute just to say wherether it conv or not
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you can assert that on the interval \((0,1)\) you have \[2xx^2>x\] so \[\sqrt{2xx^2}>\sqrt{x}\] and therefore \[\frac{1}{\sqrt{2xx^2}}<\frac{1}{\sqrt{x}}\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
since \[\int_0^1\frac{dx}{\sqrt{x}}=2\] your integral is less than or equal to 2
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i am pretty sure that is what you want to do, because the problem was cooked up so that on \((0,1)\) you have \(2xx^2>x\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[2xx^2>x\] \[xx^2>0\] \[x(1x)>0\] \[0<x<1\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
so the intergrand becomes infinite when x=0 and x=2 but 2 is of no importance since its not in the interval
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
wolfram tells me the integral is in fact \(\frac{\pi}{2}\) which is certainly less than 2, but the closed form is a big fat mess and i can't imagine trying to find it
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
I thought that this problem relates to the integral test. It should be possible to compute it, or did I mess up with the algebra?
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
integral test is usually used to determine if an infinite series converges, and you use \[\int_a^{\infty}f(x)dx\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
this is a comparison test, compare the improper integral to another improper integral that you know converges
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
yes this is what the book says thank you bot @Spacelimbus and @satellite73
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Oh I see that now @satellite73 , I just thought because by computing it you also get the value of 90° or pi/2. But I never encountered comparing the integrals :) Interesting problem anyway.
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i.e. since \(\frac{1}{\sqrt{2xx^2}}<\frac{1}{\sqrt{x}}\) you have \[\int_0^1\frac{dx}{\sqrt{2xx^2}}\leq\int_0^1\frac{dx}{\sqrt{x}}=2\]
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
@Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
somehow it gives a quarter of the circle, but i don't see it
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
well i see a primitive of the form\[ \sin^{1}x\]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Well, as I said, I was just naively computing it  I am pretty sure however that your way is more accurate. \[\Large \int \frac{1}{\sqrt{1(x1)^2}}dx \] because: \[\Large x^2+2x=(x1)^2+1 \] Substitute u=sinalpha and then you get a simple integral \[\Large \int 1 d\alpha \]
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
\[\Large \left.\sin^{1}(x1)\right^1_0 \]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
interesting
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
what if you set x1=sin theta
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
\(x1=\sin\theta\)
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
that's what I just did @electrokid
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
it'd lead to the same result as @spacelimbus
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
@Spacelimbus lol yeah
 one year ago

electrokid Group TitleBest ResponseYou've already chosen the best response.1
our brains were on parallel dimension there!
 one year ago

Spacelimbus Group TitleBest ResponseYou've already chosen the best response.2
Hehe, yes, I believe we got the same idea.
 one year ago
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