Question

Sa se demonstreze inegalitatea:
$\int\limits^a_2 {xarctgx} \, dx$ > $\int\limits^a_2 {ln(1+x^2)} \, dx$
a este 10

Am scris asa
f:[2,10]->R
f(x)=xarctgx-ln(1+x^2)
f'(x)=arctgx-$\frac{x}{x^2+1}$
f''(x)=$\frac{2x^2}{(x^2+1)^2}$ >0, pricare x ∈ [2,10]
si acum ?

Answer 1

$\displaystyle Putem~extinde~domeniul~lui~f~la~[0,10].\\ \\Continuare: \\ \\f'' \ge 0 \Rightarrow f' -crescatoare \Rightarrow f'(x) \ge f'(0)=0~\forall~x \in [0,10]. \\ \\ Deci~f' \ge 0. ~ Rezulta~f-crescatoare,~deci~f(x) \ge f(0)~\forall~x \in [0,10]. \\ \\ f(0)=0,~deci~f(x) \ge 0 ~\forall~x \in [0,10]. \\ \\ In~particular~f(x) \ge 0~\forall~x \in [2,10],~q.e.d.$