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

which one is/are right

Options:: $\displaystyle (a)\; S\geq \frac{1}{e}\;\;\;\;\;\;(b)\;S\geq 1-\frac{1}{e}\;\;\;\;\;\;(c)\;S\leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)\;\;\;\;\;\;(d)\; S\leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

$\bf{My\; Try::}$Area enclosed by $y=e^{-x^2}$ and $x=0,y=0$ and $x=1$ is

enter image description here

Given $0\leq x<1\;,$ Then $x^2\leq x,x\in \left[0,1\right)$. So $\displaystyle -x^2\geq -x ,x\in \left[0,1\right)$

So $\displaystyle \int_{0}^{1}e^{-x^2}dx \geq \int_{0}^{1}e^{-x}dx = \left(1-\frac{1}{e}\right)$. So $\displaystyle S\geq \left(1-\frac{1}{e}\right)$

Now I did not understand how can i find upper Bound of $S$

please explain me in detail

If it is possible to draw diagram, the plz explain it to me using Diagram.

Thanks

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juantheron
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1 Answers1

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This is one of the problems which appeared in the I.I.T - J.E.E exam . You can find a nicely explained solution at this link. See Question 12

C.S.
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