If S be the area of the region enclosed by $y=e^{-x^2},y=0,x=0$ and $x=1.$ Then,  which of the following option(s)  is/are correct?

Since,  $x^2\le x$ when $x\in [0,1]$

$$\begin{gathered} \Rightarrow -x^2\ge -x\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}}{e}^{-x^2}\ge e^{-x} \\ \therefore {\int }_0^1{e}^{-x^2}dx\ge {\int }_0^1{e}^{-x}dx \\ \Rightarrow S\ge -{\left(e^{-x}\right)}_0^1=1-\frac{1}{e} \mathrm{...}\left(i\right) \end{gathered}$$

Also,  ${\int }_0^1{e}^{-x^2}dx\le$ Area of two rectangles

$\begin{array}{l}\le (1\times \frac{1}{\sqrt{2}})+(1-\frac{1}{\sqrt{2}})\times \frac{1}{\sqrt{e}}\\ \le \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}}) \mathrm{...}\left(ii\right)\\ \therefore \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})\ge S\ge 1-\frac{1}{e}\end{array}$

[from Eqs.(i) and (ii)]