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. $S\le \frac{1}{4}(1+\frac{1}{\sqrt{e}})$
• D. $S\le \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$

Answer 1

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

$S\ge 1-\frac{1}{e}$
$S\ge \frac{1}{e}$
$S\le \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$

$S>\frac{1}{e}$ (As area of rectangle $OCDS=\frac{1}{e})$
Since $e^{-x^2}⩾e^{-x}\forall xϵ[0,1]$.
$\Rightarrow S>{\int }_0^1{e}^{-x}dx=(1-\frac{1}{e})$
Area of rectangle OAPQ + Area of rectangle
QBRS > S
$\Rightarrow S<\frac{1}{\sqrt{2}}\left(1\right)+(1-\frac{1}{\sqrt{2}})\left(\frac{1}{\sqrt{e}}\right)$
Since $\frac{1}{4}(1+\frac{1}{e})<1-\frac{1}{e}$