Let $S$ be the area of the region enclosed by $y=e^{-x^{2}}$, $y=0$, $x=0$ and $x=1$; then

Let S be the area of the region enclosed by y=e^{-x²}, y=0, x=0 and x=1; then $S = \int_{0}^{1} e^{-x^2}\,dx$ $0 \le x \le 1 \Rightarrow 0 \le x^2 \le 1$ $\text{To find: which of the given inequalities hold}$ $S = \int_{0}^{1} e^{-x^2}\,dx$ $x\in[0,1]:\ e^{-x^2}\ge e^{-1} \Rightarrow S \ge \int_0^1 e^{-1}\,dx = \dfrac{1}{e} \Rightarrow (1)\ \checkmark$ $e^{-u}\ge 1-u \Rightarrow e^{-x^2}\ge 1-x^2 \Rightarrow S \ge \int_0^1(1-x^2)\,dx = \dfrac{2}{3}$ $\dfrac{2}{3} \ge 1-\dfrac{1}{e} \Rightarrow (2)\ \checkmark$ $S \ge \dfrac{2}{3} \approx 0.667 > \dfrac{1}{4}\left(1+\dfrac{1}{\sqrt e}\right) \approx 0.402 \Rightarrow (3)\ \times$ $S = \int_0^{1/\sqrt2} e^{-x^2}\,dx + \int_{1/\sqrt2}^{1} e^{-x^2}\,dx$ $\le \dfrac{1}{\sqrt2}\cdot 1 + \left(1-\dfrac{1}{\sqrt2}\right)e^{-1/2} = \dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt e}\left(1-\dfrac{1}{\sqrt2}\right) \Rightarrow (4)\ \checkmark$ $\boxed{\text{Correct options: } (1),\ (2),\ (4)}$