[tex]\displaystyle g(x) =\int_{0}^{x^2} e^{t^2}\, dt \\\\ \text{Fie }F(t)\text{ primitiva functiei }f(t) = e^{t^2}\\ \\\\ g(x) =\int_{0}^{x^2} e^{t^2}\, dt= F(t)\Big|_{0}^{x^2} = F(x^2)-F(0) \\ \\ \\\text{Derivand si aplicand regula lantului obtinem:}\\ \\ g'(x) = \Big(F(x^2)\Big)' - \Big(F(0)\Big)' = F'(x^2)\cdot (x^2)' - 0 =\\ \\= f(x^2)\cdot (2x) =e^{x^4}\cdot (2x) = 2x\cdot e^{x^4}\\ \\\\ g''(x) =2e^{x^4}+8x^4e^{x^4} \Rightarrow g''(x) > 0,\quad \forall x\in \mathbb{R}[/tex]
⇒ d) g este convexă.
