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Test for convergence

Mathematics
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\[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\]
improper intergral
Not sure how much that helps but: \[\Large -x^2+2x=-(x^2-2x-1+1)=-(x-1)^2+1 \]

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Other answers:

\[\Large \int\frac{dx}{\sqrt{1-(x-1)^2}} \]
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
yes we dont really want to compute just to say wherether it conv or not
you can assert that on the interval \((0,1)\) you have \[2x-x^2>x\] so \[\sqrt{2x-x^2}>\sqrt{x}\] and therefore \[\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\]
since \[\int_0^1\frac{dx}{\sqrt{x}}=2\] your integral is less than or equal to 2
i am pretty sure that is what you want to do, because the problem was cooked up so that on \((0,1)\) you have \(2x-x^2>x\)
\[2x-x^2>x\] \[x-x^2>0\] \[x(1-x)>0\] \[0
so the intergrand becomes infinite when x=0 and x=2 but 2 is of no importance since its not in the interval
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
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?
integral test is usually used to determine if an infinite series converges, and you use \[\int_a^{\infty}f(x)dx\]
this is a comparison test, compare the improper integral to another improper integral that you know converges
yes this is what the book says thank you bot @Spacelimbus and @satellite73
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.
i.e. since \(\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\) you have \[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\leq\int_0^1\frac{dx}{\sqrt{x}}=2\]
@Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me
somehow it gives a quarter of the circle, but i don't see it
well i see a primitive of the form\[ \sin^{-1}x\]
Well, as I said, I was just naively computing it - I am pretty sure however that your way is more accurate. \[\Large \int \frac{1}{\sqrt{1-(x-1)^2}}dx \] because: \[\Large -x^2+2x=-(x-1)^2+1 \] Substitute u=sinalpha and then you get a simple integral \[\Large \int 1 d\alpha \]
\[\Large \left.\sin^{-1}(x-1)\right|^1_0 \]
interesting
what if you set x-1=sin theta
\(x-1=\sin\theta\)
that's what I just did @electrokid
it'd lead to the same result as @spacelimbus
@Spacelimbus lol yeah
our brains were on parallel dimension there!
Hehe, yes, I believe we got the same idea.

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