# How to explicit the $\min$ function in integral?

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!

• 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? – James Kilbane Feb 20 '14 at 10:19

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 : $$\frac{x^2}{2}|_0^1 + 2 \tan^{-1} x |_1^2 = \frac{1}{2} + 2 ( \tan^{-1} 2 - \frac{\pi}{4} ).$$
• @DonAntonio. The solution is for $x=2/(1+x^2)$ and not $x=1/(2+x^2)$. Haha ! Muy amistosamente. – Claude Leibovici Feb 20 '14 at 10:55