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I have a function $f:[0,2]\to \mathbb R$ and $f(x)=\min \bigg (x,\cfrac{2}{1+x^2} \bigg)$. How do I explicit this function to solve the following integral: $$\int_{0}^{2}{f(x)dx}$$

Thank you!

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    $\begingroup$ Hint, where is x > 2/(1+x^2)? Where is 2/(1+x^2) > x? Can you split the integral up onto the two domains? $\endgroup$ – James Kilbane Feb 20 '14 at 10:19
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For $ x = \frac{2}{1+x^2} $, one of the roots is 1 and other two are complex. Because one is a str. line and another is decreasing curve.

So separate the integral into two parts according to which is minimum where : $$ \int_{0}^{1} x dx + \int_{1}^{2} \frac{2}{1+x^2} dx $$ and there you are done.

Value of integral : $$ \frac{x^2}{2}|_0^1 + 2 \tan^{-1} x |_1^2 = \frac{1}{2} + 2 ( \tan^{-1} 2 - \frac{\pi}{4} ). $$

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    $\begingroup$ @DonAntonio. The solution is for $x=2/(1+x^2)$ and not $x=1/(2+x^2)$. Haha ! Muy amistosamente. $\endgroup$ – Claude Leibovici Feb 20 '14 at 10:55
  • $\begingroup$ @ClaudeLeibovici : typing mistake, edited ! thanks. :) $\endgroup$ – DiffeoR Feb 20 '14 at 10:56

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