Question

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

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

Answer 1

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

$S>\frac{1}{e}$ (As area of rectangle $OCDS=1/e)$
Since $e^{-x^2}\ge e^{-x}\forall x\in [0,1]$
Since $e^{-x^2}\ge e^{-x}\forall x\in [0,1]$
$\Rightarrow S>\underset{0}{\overset{1}{\int }}{e}^{-x}dx=\left(1-\frac{1}{e}\right)$
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 $\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)<1-\frac{1}{e}$