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

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Aicharya · Accepted Answer

The given curve $y=e^{-x^{2}}$ Draw a rough sketch of curve at $x=0, y=1$ and at $x=1, y=1 / e$ $\because y=e^{-x^{2}}$ $\Rightarrow \frac{d y}{d x}=-2 x e^{-x^{2}} < 0 \quad \forall x \in(0,1)$ $\therefore y=e^{-x^{2}}$ is decreasing on $(0,1)$ Hence its graph is as shown in figure given below

Now, $\mathrm{S}=$ area exclosed by curve $=\mathrm{XYCO}$ and area of rectangle $\mathrm{OCYL}=\frac{1}{e}$ Clearly $S>\frac{1}{e} \quad \therefore$ a is true. For $x \in[0,1] \Rightarrow x^{2} < x$ $\begin{array}{l} \Rightarrow-x^{2}>-x \quad \Rightarrow e^{-x^{2}} \geq e^{-x} \forall x \in[0,1] \\ \Rightarrow \int_{0}^{1} e^{-x^{2}} d x>\int_{0}^{1} e^{-x} d x=1-\frac{1}{e} \\ \Rightarrow S>1-\frac{1}{e} \quad \therefore \text { (b) is true. } \end{array}$ Now $S < $ area of rectangle $\mathrm{XADO}+$ area of rectangle $\mathrm{ZDCN}$ $\Rightarrow S < \frac{1}{\sqrt{2}} \times 1+\left(1-\frac{1}{\sqrt{2}}\right) \frac{1}{\sqrt{e}}$ $\therefore S < \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right) \quad \because(\mathrm{d})$ is true Also as $\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right) < 1-\frac{1}{e} \quad \therefore$ (c) is incorrect.

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

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