$$\begin{array}{lc} & \mathrm{S}=\int_0^1 e^{-x^2} d x \\ \text { Now, } & -x^2 \leq 0 \\ \Rightarrow & e^{-x^2} \leq 1 \\ \Rightarrow & \int_0^1 e^{-x^2} d x \leq 1 \\ \text { Now, } & x<1 \\ \Rightarrow & x^2 \leq x \\ \Rightarrow & -x^2 \geq-x \\ \Rightarrow & e^{-x^2} \geq e^{-x} \\ \Rightarrow & \mathrm{S} \geq \int_0^1 e^{-x} d x \end{array}$$

$$\begin{array}{ll} \Rightarrow & S \geq-\left(e^{-x}\right)_0^1 \\ \Rightarrow & S \geq-\left(\frac{1}{e}-1\right) \\ \Rightarrow & S \geq 1-\frac{1}{e} \Rightarrow(B) \text { is correct. } \\ \text { Since, } & S \geq 1-\frac{1}{e} \\ \Rightarrow & S>\frac{1}{e} \Rightarrow(\mathrm{A}) \text { is correct. } \end{array}$$

Now, Area of rectangle OAPQ + Area of rectangle QBRS > S

$$\Rightarrow \mathrm{S}<\frac{1}{\sqrt{2}}(1)+\left(1-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{e}}\right) \Rightarrow(\mathrm{D})$$ is correct.

Since, $$\quad \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)<1-\frac{1}{e}$$

Hence, $$(\mathrm{C})$$ is incorrect.