anonymous
  • anonymous
Test for convergence
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\]
anonymous
  • anonymous
improper intergral
anonymous
  • anonymous
Not sure how much that helps but: \[\Large -x^2+2x=-(x^2-2x-1+1)=-(x-1)^2+1 \]

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

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