Area Question 29
Question 29
- If $S$ be the area of the region enclosed by $y=e^{-x^{2}}, y=0, x=0$ and $x=1$. Then,
(2012) (a) $S \geq \frac{1}{e}$ (b) $S \geq 1-\frac{1}{e}$ (c) $S \leq \frac{1}{4} 1+\frac{1}{\sqrt{e}}$ (d) $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}} 1-\frac{1}{\sqrt{2}}$
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Solution:
- PLAN (i) Area of region $f(x)$ bounded between $x=a$ to $x=b$ is
$\int_{a}^{b} f(x) d x=$ Sum of areas of rectangle shown in shaded part.
(ii) If $f(x) \geq g(x)$ when defined in $[a, b]$, then
$$ \int_{a}^{b} f(x) d x \geq \int_{a}^{b} g(x) d x $$
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, $x^{2} \leq x$ when $x \in[0,1]$
$$ \begin{array}{lrlrl} & \Rightarrow & -x^{2} & \geq-x & \text { or } e^{-x^{2}} \geq e^{-x} \ & \therefore & & \int_{0}^{1} e^{-x^{2}} d x & \geq \int_{0}^{1} e^{-x} d x \ & \Rightarrow & & S & \geq-\left(e^{-x}\right)_{0}^{1}=1-\frac{1}{e} \end{array} $$
Also, $\int_{0}^{1} e^{-x^{2}} d x \leq$ Area of two rectangles
$$ \begin{aligned} & \leq 1 \times \frac{1}{\sqrt{2}}+1-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{e}} \ & \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}} 1-\frac{1}{\sqrt{2}} \end{aligned} $$
$\therefore \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}} 1-\frac{1}{\sqrt{2}} \geq S \geq 1-\frac{1}{e} \quad$ [from Eqs. (i) and (ii)]
