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

improper intergral

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

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

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

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

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

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

@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\[ \sin^{-1}x\]

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

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.