Area (1 Viewer)

juantheron · Apr 8, 2012

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:

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

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

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

$\displaystyle (d)\;\;\mathbb{S}\leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

juantheron · Apr 8, 2012

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,Help required...

Thanks

math man · Apr 8, 2012

bleakarcher · Apr 8, 2012

juantheron said:
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,Help required...

Thanks

the statement i bolded is wrong since you are talking about the interval [0,1].

EDIT:
ignore the above, your statement is correct. sorry, was tired last night

jyu · Apr 9, 2012

Both (a) and (b) are right. stupid choices

math man · Apr 9, 2012

(a) is more corrct though from my diagram.

seanieg89 · Apr 11, 2012

Options a, b and d are all correct. And none is "more correct" than any other. Perhaps the question asked you to single out the incorrect statement?

seanieg89 · Apr 11, 2012

(To prove the upper bound in d, bound the integrand above separately on the intervals [0,1/sqrt(2)] and [1/sqrt(2),1].)

Area (1 Viewer)

juantheron

Active Member

juantheron

Active Member

math man

Member

bleakarcher

Active Member

jyu

Member

math man

Member

seanieg89

Well-Known Member

seanieg89

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)