Question

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

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

Answer 1

The correct option is C

$\mathrm{S}\le \frac{1}{4}\left(1+\frac{1}{\sqrt{\mathrm{e}}}\right)$

Explanation for the correct option.

Step 1: Graph

Given a region is bounded by: $\mathrm{y}={\mathrm{e}}^{-{\mathrm{x}}^2},\mathrm{y}=0,\mathrm{x}=0$, and $\mathrm{x}=1$.

Step 2: Integrate the function

As area of rectangle $\mathrm{OCDS}=\frac{1}{\mathrm{e}}$.

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

Hence, option C is correct.