[tex]\min\Big\{x,\dfrac{2}{1+x^2}\Big\} = \begin{cases} x,\ x\in (0;\,1) \\\\ \dfrac{2}{1+x^2},\, x\in(1;\,2)\end{cases}\\ \\ \text{E clar de ce, fiindca daca: } \\ x\in (0;\,1)\Rightarrow x\in (0;1),\quad \dfrac{2}{1+x^2}\in (1;\,2)\\ x\in(1;\,2) \Rightarrow x \in (1;\,2),\quad \dfrac{2}{1+x^2} \in (0,..;\, 1)\\ \\ \displaystyle \int_{0}^2\min\Big\{x,\dfrac{2}{1+x^2}\Big\} \, dx= \int_{0}^1 x\, dx+\int_1 ^2 \dfrac{2}{1+x^2}\, dx= \\ \\ = \dfrac{1}{2}+2\arctan(x)\Big|_{1}^2 = \dfrac{1}{2}+2\arctan 2 - \dfrac{\pi}{2}[/tex]
Răspuns corect b).
