Question

Let S be the area of the region enclosed by $y=e^{-x^2},y=0,x=0,x=1.$ Then,
A) $S⩾{\displaystyle \frac{1}{e}}$
B) $S⩾1-{\displaystyle \frac{1}{e}}$
C) $S⩽{\displaystyle \frac{1}{4}}(1+{\displaystyle \frac{1}{\sqrt{e}}})$
D) $S⩽{\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{e}}}(1-{\displaystyle \frac{1}{\sqrt{2}}})$

Answer 1

Hint: First we have to draw the figure and understand the area $S$. Then the first two options can be checked simply using the values of $x,y$. Using B, C can be checked. Certain rearrangements are needed to check D. The concept used here is that the area under a curve is equal to the integral value. For applying this we have to find the limits as well.

Formula used:
Area enclosed by a curve $y=f(x)$ within the limits $x=a$ to $x=b$ is simply the integral of the curve with same limits.
$\Rightarrow S=\underset{a}{\overset{b}{\int }}f(x)dx$

Complete step by step answer:
Given that $S$ is the area enclosed by $y=e^{-x^2},y=0,x=0,x=1.$
$y=0,x=0$ represent the $x,y$ axis respectively.
Let the curve drawn represent $y=e^{-x^2}$.

In the figure, the shaded portion represents the area $S$.
When $x=1\Rightarrow y=e^{-x^2}=e^{-1}={\displaystyle \frac{1}{e}}$.
Since $S$ contains a rectangle OSDC with vertices $(0,0),(1,0),(1,{\displaystyle \frac{1}{e}}),(0,{\displaystyle \frac{1}{e}})$, where S(1,0).
The area of that rectangle will be $A=length\times breadth$
$\Rightarrow A=1\times {\displaystyle \frac{1}{e}}={\displaystyle \frac{1}{e}}$.
Therefore clearly $S⩾{\displaystyle \frac{1}{e}}$.
This implies option A is right.
Also, since $S$ is the area enclosed by the curve $y=e^{-x^2},x=0,y=0,x=1$
We can consider $S=\underset{0}{\overset{1}{\int }}{e}^{-x^2}dx$.
$\Rightarrow x\in [0,1]\Rightarrow x^2⩽x$
Multiplying both sides by $-1$ and taking exponents we have,
$\Rightarrow -x^2⩾-x\Rightarrow e^{-x^2}⩾e^{-x}$
Therefore, $S=\underset{0}{\overset{1}{\int }}{e}^{-x^2}dx⩾\underset{0}{\overset{1}{\int }}{e}^{-x}dx$
Integrating we get, $S⩾[-e^{-x}{]}_0^1=e^0-e^{-1}=1-{\displaystyle \frac{1}{e}}$
$\Rightarrow S⩾1-{\displaystyle \frac{1}{e}}$
Therefore option B is also right.
To show that option C is incorrect,
$\Rightarrow e>\sqrt{e}\Rightarrow {\displaystyle \frac{1}{e}}<{\displaystyle \frac{1}{\sqrt{e}}}$
$\Rightarrow -{\displaystyle \frac{1}{e}}<{\displaystyle \frac{1}{\sqrt{e}}}\Rightarrow 1-{\displaystyle \frac{1}{e}}<1+{\displaystyle \frac{1}{\sqrt{e}}}<{\displaystyle \frac{1}{4}}(1+{\displaystyle \frac{1}{\sqrt{e}}})$
This implies $1-{\displaystyle \frac{1}{e}}<{\displaystyle \frac{1}{4}}(1+{\displaystyle \frac{1}{\sqrt{e}}})$
From B, we have $S⩾1-{\displaystyle \frac{1}{e}}$
Using these two equations we can say C is incorrect.
Moving on, consider two rectangles in the figure OAPQ and QBRS where $S(1,0)$,
Clearly $S⩽$ Area of OAPQ$+$ Area of QBRS
PQ is drawn parallel to Y axis at the point $x={\displaystyle \frac{1}{\sqrt{2}}}$.
Then $y=e^{-x^2}=e^{-{\displaystyle \frac{1}{2}}}={\displaystyle \frac{1}{\sqrt{e}}}$.
Therefore, B is the point $({\displaystyle \frac{1}{\sqrt{2}}},{\displaystyle \frac{1}{\sqrt{e}}})$.
Area of OAPQ $=lenth\times breadth=1\times {\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{1}{\sqrt{2}}}$
In QBRS, $QB={\displaystyle \frac{1}{\sqrt{e}}},BR=1-{\displaystyle \frac{1}{\sqrt{2}}}$
Therefore, Area of QBRS $=length\times breadth={\displaystyle \frac{1}{\sqrt{e}}}\times 1-{\displaystyle \frac{1}{\sqrt{2}}}$
$\Rightarrow S⩽AreaOAPQ+AreaQBRS={\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{e}}}(1-{\displaystyle \frac{1}{\sqrt{2}}})$
So, option D is correct.

Therefore, Option (A), (B) and (D) are correct.

Note:
It is important to identify the graph of the given function and mark the given region. If there is some portion under the X-axis, this area will be negative. So be careful in those cases to split the region if necessary and find the area separately.