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anonymous
 one year ago
Find the value of the integral !
anonymous
 one year ago
Find the value of the integral !

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{2}\sqrt{\frac{ 2x }{ x1 }}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Studied the convergence and found that it converges

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now I must find to what it converges

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[u=\sqrt{x1} \\ u^2=x1 \\ 2u du=dx \\ \int\limits_0^1 \frac{\sqrt{2(u^2+1)}}{u} 2 u du \\ 2 \int\limits_0^1 \sqrt{1u^2} du\] try this integral

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you actually can find that using just a geometrical approach it is just a quarter of a circle with radius 1 multiplied by 2

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i'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 \( \lim_{a \rightarrow 1+} \ \int_{a}^{2} \sqrt{\frac{2x}{x1}}\). no idea if this is helpful, but hope it does help.
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