SATHEE: Area Question 29

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) $S \leq \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}} 1 - \frac{1}{\sqrt{2}}$

Show Answer

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 $x \in [0, 1]$

$\begin{array}{lrlrl} \Rightarrow & - x^{2} & \geq - x & or e^{- x^{2}} \geq e^{- x} & ∴ & \int_{0}^{1} e^{- x^{2}} d x & \geq \int_{0}^{1} e^{- x} d x & \Rightarrow & S & \geq - {(e^{- x})}_{0}^{1} = 1 - \frac{1}{e} \end{array}$

Also, $\int_{0}^{1} e^{- x^{2}} d x \leq$ Area of two rectangles

$\begin{aligned} \leq 1 \times \frac{1}{\sqrt{2}} + 1 - \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{e}} & \leq \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}} 1 - \frac{1}{\sqrt{2}} \end{aligned}$

$∴ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}} 1 - \frac{1}{\sqrt{2}} \geq S \geq 1 - \frac{1}{e}$ [from Eqs. (i) and (ii)]

Area Question 29

Question 29

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Area Question 29

Question 29

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu