Question

Find the value of the integral !

Answer 1

\[\int\limits_{1}^{2}\sqrt{\frac{ 2-x }{ x-1 }}\]

Answer 2

Studied the convergence and found that it converges

Answer 3

now I must find to what it converges

Answer 4

\[u=\sqrt{x-1} \\ u^2=x-1 \\ 2u du=dx \\ \int\limits_0^1 \frac{\sqrt{2-(u^2+1)}}{u} 2 u du \\ 2 \int\limits_0^1 \sqrt{1-u^2} du\]
try this integral

Answer 5

you actually can find that using just a geometrical approach
it is just a quarter of a circle with radius 1 multiplied by 2

Answer 6

i've run a simple numerical on this and it tells me that the answer is 1 but that you need to be very careful about the start of the interval [which is why i ran the numerical in first place]. this makes me wonder if this is really to be approached as \( \lim_{a \rightarrow 1+} \ \int_{a}^{2} \sqrt{\frac{2-x}{x-1}}\).

no idea if this is helpful, but hope it does help.