- anonymous

Test for convergence

- katieb

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- anonymous

\[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\]

- anonymous

improper intergral

- 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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## More answers

- anonymous

\[\Large \int\frac{dx}{\sqrt{1-(x-1)^2}} \]

- 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

yes we dont really want to compute just to say wherether it conv or not

- 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

since \[\int_0^1\frac{dx}{\sqrt{x}}=2\]
your integral is less than or equal to 2

- 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

\[2x-x^2>x\]
\[x-x^2>0\]
\[x(1-x)>0\]
\[0

- 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

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

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

integral test is usually used to determine if an infinite series converges, and you use
\[\int_a^{\infty}f(x)dx\]

- anonymous

this is a comparison test, compare the improper integral to another improper integral that you know converges

- anonymous

yes this is what the book says thank you bot @Spacelimbus and @satellite73

- 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

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

@Spacelimbus it it clear why it is \(\frac{\pi}{2}\) ? it is not clear to me

- anonymous

somehow it gives a quarter of the circle, but i don't see it

- anonymous

well i see a primitive of the form\[ \sin^{-1}x\]

- 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

\[\Large \left.\sin^{-1}(x-1)\right|^1_0 \]

- anonymous

interesting

- anonymous

what if you set x-1=sin theta

- anonymous

\(x-1=\sin\theta\)

- anonymous

that's what I just did @electrokid

- anonymous

it'd lead to the same result as @spacelimbus

- anonymous

@Spacelimbus lol yeah

- anonymous

our brains were on parallel dimension there!

- anonymous

Hehe, yes, I believe we got the same idea.

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