Question

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

• A. $ (a) S \ge \frac{1}{e} $
• B. $ (b) S \ge 1 - \frac{1}{e} $
• C. $ (c) S \le \frac{1}{4} \bigg (1 + \frac{1}{\sqrt e}\bigg)$
• D. $ (d) S \le \frac{1}{\sqrt 2} + \frac{1}{\sqrt e} \bigg (1 - \frac{1}{\sqrt 2}\bigg)$

Answer 1

The Correct Option is D

(i) Area of region f(x) bounded between x =a to x = b is 
\(\, \, \, \, \int \limits_a^b f(x) dx\) =Sum of areas of rectangle shown in shaded part 
(ii) If f{x)>g(x) when defined in [a,b], then 
\(\, \, \, \, \, \int \limits_a^b f(x) dx \ge \int \limits_a^b g(x) dx\) 
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. 
Graph for \(y = e^{-x^2}\) 
Since, \(x^2 \le x\, \, when\, \, x \in [0 , 1]\)
\(\Rightarrow - x^2 \ge -x\, \, or\, \, e^{-x^2} \ge e^{-x}\)
\(\therefore \, \, \, \int \limits_0^1 e^{-x^2} dx \ge \int \limits_0^1 e^{-x} dx\)
\(\Rightarrow \, \, \, \, \, \, S \ge -(e^{-x})_0^1 = 1 - \frac{1}{e} \, \, \, .........(i)\) 
Also, \(\int \limits_0^1 e^{-x^2} dx \le\) Area of two rectangles 
\(\, \, \, \, \, \, \, \, \, \, \le \bigg(1 \times \frac{1}{\sqrt2}\bigg) + \bigg(1 - \frac{1}{\sqrt2}\bigg) \times \frac{1}{\sqrt e}\)
\(\, \, \, \, \, \, \, \, \le \frac{1}{\sqrt2} + \frac{1}{\sqrt e} \bigg(1 - \frac{1}{\sqrt2}\bigg)\, \, \, \, ....(ii)\)
\(\therefore \frac{1}{\sqrt2} + \frac {1}{\sqrt e}\bigg(1 - \frac{1}{\sqrt2}\bigg) \ge\, \, S\, \, \ge 1 - \frac{1}{e}\, \, \, [from Eqs. (i) and (ii)]\)

Answer 2

Given that;
\(y=e^{-x^{2}},y=0,x=0,x=1\)
and ‘S’ be the area enclosed by above
\(\therefore\) i.e.,\(S≥\frac{1}{e}\)
Since area of rectangle\(=\frac{1}{e}\)
\(=e^{-x^{2}}\geq e^{-x}\forall x\in[0,1]\)
\(\therefore S\geq \int^1_0e^{-x}dx=(1-\frac{1}{e})\)
Area of rectangle \(QBRS\gt S\)
\(=S\leq\frac{1}{\sqrt{2}}(1)+(1-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{e}})\)
\(\therefore \frac{1}{4}(1+\frac{1}{\sqrt{e}})\leq(1-\frac{1}{e})\)