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Jonask
 one year ago
Test for convergence
Jonask
 one year ago
Test for convergence

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i 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

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

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

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

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1wolfram 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

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

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

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

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

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.2Oh 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.

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

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

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

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

Spacelimbus
 one year ago
Best ResponseYou've already chosen the best response.2Well, 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 \[\Large \int 1 d\alpha \]

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

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

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

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

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

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

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