Question

Test for convergence

Answer 1

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

Answer 2

improper intergral

Answer 3

Not sure how much that helps but:
\[\Large -x^2+2x=-(x^2-2x-1+1)=-(x-1)^2+1 \]

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

you can assert that on the interval \((0,1)\) you have
\[2x-x^2>x\] so
\[\sqrt{2x-x^2}>\sqrt{x}\] and therefore
\[\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\]

Answer 8

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

Answer 9

i am pretty sure that is what you want to do, because the problem was cooked up so that on \((0,1)\) you have \(2x-x^2>x\)

Answer 10

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

Answer 11

so the intergrand becomes infinite when x=0 and x=2
but 2 is of no importance since its not in the interval

Answer 12

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

Answer 13

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?

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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.

Answer 18

i.e. since \(\frac{1}{\sqrt{2x-x^2}}<\frac{1}{\sqrt{x}}\) you have
\[\int_0^1\frac{dx}{\sqrt{2x-x^2}}\leq\int_0^1\frac{dx}{\sqrt{x}}=2\]

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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
\[\Large \int 1 d\alpha \]

Answer 23

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

Answer 24

interesting

Answer 25

what if you set x-1=sin theta

Answer 26

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

Answer 27

that's what I just did @electrokid

Answer 28

it'd lead to the same result as @spacelimbus

Answer 29

@Spacelimbus lol yeah

Answer 30

our brains were on parallel dimension there!

Answer 31

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