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anonymous

  • one year ago

Find the value of the integral !

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  1. anonymous
    • one year ago
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    \[\int\limits_{1}^{2}\sqrt{\frac{ 2-x }{ x-1 }}\]

  2. anonymous
    • one year ago
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    Studied the convergence and found that it converges

  3. anonymous
    • one year ago
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    now I must find to what it converges

  4. freckles
    • one year ago
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    \[u=\sqrt{x-1} \\ u^2=x-1 \\ 2u du=dx \\ \int\limits_0^1 \frac{\sqrt{2-(u^2+1)}}{u} 2 u du \\ 2 \int\limits_0^1 \sqrt{1-u^2} du\] try this integral

  5. freckles
    • one year ago
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    you actually can find that using just a geometrical approach it is just a quarter of a circle with radius 1 multiplied by 2

  6. IrishBoy123
    • one year ago
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    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.

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