Answer:
Option A,B,D
Explanation:
Concept involved
(i) Area of region f(x) bounded between x=a to x=b is
$\int_{a}^{b} f(x) dx=$ sum of areas of rectangle shown in shaded part
(ii) if $f(x) \geq g(x) $ when defined in [a, b]
$\Rightarrow$ $\int_{a}^{b} f(x) dx \geq\int_{a}^{b} g(x) dx$
Description of Situation As the given curve y =$e^{-x^{2}}$ cannot be integrated thus we
have to bound this function by using above mentioned concept
sol. Graph for , $y=e^{-x^{2}}$
since , $x^{2} \leq x$ when $x \epsilon [0,1]$
$\Rightarrow$ $-x^{2} \geq -x$ or $e^{-x^{2}} \geq e^{-x}$
$\therefore$ $\int_{0}^{1} e^{-x^{2}} dx \geq \int_{0}^{1} e^{-x} dx$
$\Rightarrow$ $S\geq -(e^{-x})_{0}^{1} =1-\frac{1}{e}$........(i)
Also, $\int_{0}^{1} e^{-x^{2}} dx \leq $ area of two rectangle
$\leq \left(1\times \frac{1}{\sqrt{2}}\right)+\left(1-\frac{1}{\sqrt{2}}\right)\times\frac{1}{\sqrt{e}}$
$\leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$..........(ii)
$\therefore$ $ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)\geq S \geq1-\frac{1}{e}$
[from Eqs.(i) and (ii) ]
