## Jonask Group Title Test for convergence one year ago one year ago

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

improper intergral

3. Spacelimbus

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

4. Spacelimbus

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

5. satellite73

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

7. satellite73

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

8. satellite73

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

9. satellite73

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

10. satellite73

$2x-x^2>x$ $x-x^2>0$ $x(1-x)>0$ $0<x<1$

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

12. satellite73

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

13. Spacelimbus

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?

14. satellite73

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

15. satellite73

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

17. Spacelimbus

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.

18. satellite73

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

19. satellite73

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

20. satellite73

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

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

22. Spacelimbus

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 $\Large \int 1 d\alpha$

23. Spacelimbus

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

interesting

25. electrokid

what if you set x-1=sin theta

26. electrokid

$$x-1=\sin\theta$$

27. Spacelimbus

that's what I just did @electrokid

28. electrokid

it'd lead to the same result as @spacelimbus

29. electrokid

@Spacelimbus lol yeah

30. electrokid

our brains were on parallel dimension there!

31. Spacelimbus

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