# area with minimum and maximum

Printable View

• April 8th 2012, 03:12 AM
jacks
area with minimum and maximum
If $\mathbb{S}$ be The Area of The Region enclosed by $y=e^{-x^2}\;,x=0\;,y=0\;,x=1$. Then

Which one is Right

options:

$\displaystyle (a)\;\;\mathbb{S}\geq \frac{1}{e}$

$\displaystyle (b)\;\;\mathbb{S}\geq 1-\frac{1}{e}$

$\displaystyle (c)\;\;\mathbb{S}\leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$

$\displaystyle (d)\;\;\mathbb{S}\leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$
• April 8th 2012, 03:13 AM
jacks
Re: area with minimum and maximum
I have Tried for lower bond

like $\displaystyle \int_{0}^{1}e^{-x^2}dx\geq \int_{0}^{1}e^{-x}dx=(1-\frac{1}{e})$

So I am getting (ii) option

How can I calculate for Upper Bond