Question

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

• A. $S \geq \frac{1}{e}$
• B. $S \geq 1-\frac{1}{e}$
• C. $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
• D. $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (A), (B), (D)

$S >\frac{1}{e} $ (As area of rectangle $\left.OCDS =1 / e \right)$
Since $e ^{- x ^{2}} \geq e ^{-x} \forall x \in[0,1]$
Since $ e ^{- x ^{2}} \geq e ^{- x } \forall x \in[0,1] $
$\Rightarrow S>\int\limits_{0}^{1} e^{-x} d x=\left(1-\frac{1}{e}\right)$
Area of rectangle OAPQ + Area of rectangle QBRS $> S$
$S<\frac{1}{\sqrt{2}}(1)+\left(1-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{e}}\right)$
Since $\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)<1-\frac{1}{e}$