Question

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

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

Answer 1

Answer: (B), (C), (D)

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

$x\in [0,1]\Rightarrow x^2\le x\Rightarrow -x^2\ge -x\Rightarrow e^{-x^2}\ge e^{-x}$
$S=\underset{0}{\overset{1}{\int }}{e}^{-x^2}dx\ge \underset{0}{\overset{1}{\int }}{e}^{-x}dx=[-e^{-x}{]}_0^1=1-\frac{1}{e}$
i.e., $S\ge 1-\frac{1}{e}\Rightarrow$ option (2) is correct.

$S$ is the area under the curve $y=e^{-x^2}$
Hence, $S\ge$ area of the rectangle $OPQS$
i.e., $S\ge 1\times \frac{1}{e}=\frac{1}{e}$
Hence, option (1) is correct.
$S=\underset{0}{\overset{1}{\int }}{e}^{-x^2}dx$
$=\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}{e}^{-x^2}dx+\underset{\frac{1}{\sqrt{2}}}{\overset{1}{\int }}{e}^{-x^2}dx$

$\le$ Area $A+$ Area $B$
Now, Area $A=\underset{0}{\overset{\frac{1}{\sqrt{2}}}{\int }}1.dx=\frac{1}{\sqrt{2}}$
Area $B=\underset{\frac{1}{\sqrt{2}}}{\overset{1}{\int }}\frac{1}{\sqrt{e}}dx=\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$
Hence, $S\le \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}(1-\frac{1}{\sqrt{2}})$
Hence, option (4) is correct
Now, consider the integral $\underset{1}{\overset{e}{\int }}\frac{dx}{x^{3/2}}=2[1-\frac{1}{\sqrt{e}}]$
Substitute $t=\ln x$
$\Rightarrow \underset{1}{\overset{e}{\int }}\frac{dx}{x^{3/2}}=\underset{0}{\overset{1}{\int }}{e}^{-t/2}dt$
Now, for $0<x<1$
$e^{-x^2}>e^{-x/2}$
$\Rightarrow \underset{0}{\overset{1}{\int }}{e}^{-x^2}dx>2[1-\frac{1}{\sqrt{2}}]$
Now, $2[1-\frac{1}{\sqrt{e}}]-\frac{1}{4}(1+\frac{1}{\sqrt{e}})=\frac{7}{4}-\frac{9}{4\sqrt{e}}$
$=\frac{7\sqrt{e}-9}{4\sqrt{e}}>0$
So, option (3) is not correct.