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anonymous
 3 years ago
Test for convergence
anonymous
 3 years ago
Test for convergence

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not sure how much that helps but: \[\Large x^2+2x=(x^22x1+1)=(x1)^2+1 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \int\frac{dx}{\sqrt{1(x1)^2}} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since \[\int_0^1\frac{dx}{\sqrt{x}}=2\] your integral is less than or equal to 2

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

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

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wolfram 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I thought that this problem relates to the integral test. It should be possible to compute it, or did I mess up with the algebra?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0integral test is usually used to determine if an infinite series converges, and you use \[\int_a^{\infty}f(x)dx\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is a comparison test, compare the improper integral to another improper integral that you know converges

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i.e. since \(\frac{1}{\sqrt{2xx^2}}<\frac{1}{\sqrt{x}}\) you have \[\int_0^1\frac{dx}{\sqrt{2xx^2}}\leq\int_0^1\frac{dx}{\sqrt{x}}=2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0somehow it gives a quarter of the circle, but i don't see it

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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 \[\Large \int 1 d\alpha \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large \left.\sin^{1}(x1)\right^1_0 \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what if you set x1=sin theta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's what I just did @electrokid

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it'd lead to the same result as @spacelimbus

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Spacelimbus lol yeah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0our brains were on parallel dimension there!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hehe, yes, I believe we got the same idea.
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