Discussion Forum

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

Answer:

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) ]