Solveeit Logo

Question

Question: Let S be the area of the region enclosed by \(y = {e^{ - {x^2}}},y = 0,x = 0,x = 1.\) Then, A) \(S...

Let S be the area of the region enclosed by y=ex2,y=0,x=0,x=1.y = {e^{ - {x^2}}},y = 0,x = 0,x = 1. Then,
A) S1eS \geqslant \dfrac{1}{e}
B) S11eS \geqslant 1 - \dfrac{1}{e}
C) S14(1+1e)S \leqslant \dfrac{1}{4}(1 + \dfrac{1}{{\sqrt e }})
D) S12+1e(112)S \leqslant \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt e }}(1 - \dfrac{1}{{\sqrt 2 }})

Explanation

Solution

First we have to draw the figure and understand the area SS. Then the first two options can be checked simply using the values of x,yx,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)y = f(x) within the limits x=ax = a to x=bx = b is simply the integral of the curve with same limits.
S=abf(x)dx\Rightarrow S = \int\limits_a^b {f(x)dx}

Complete step by step answer:
Given that SS is the area enclosed by y=ex2,y=0,x=0,x=1.y = {e^{ - {x^2}}},y = 0,x = 0,x = 1.
y=0,x=0y = 0,x = 0 represent the x,yx,y axis respectively.
Let the curve drawn represent y=ex2y = {e^{ - {x^2}}}.

In the figure, the shaded portion represents the area SS.
When x=1y=ex2=e1=1ex = 1 \Rightarrow y = {e^{ - {x^2}}} = {e^{ - 1}} = \dfrac{1}{e}.
Since SS contains a rectangle OSDC with vertices (0,0),(1,0),(1,1e),(0,1e)(0,0),(1,0),(1,\dfrac{1}{e}),(0,\dfrac{1}{e}), where S(1,0).
The area of that rectangle will be A=length×breadthA = length \times breadth
A=1×1e=1e\Rightarrow A= 1 \times \dfrac{1}{e} = \dfrac{1}{e}.
Therefore clearly S1eS \geqslant \dfrac{1}{e}.
This implies option A is right.
Also, since SS is the area enclosed by the curve y=ex2,x=0,y=0,x=1y = {e^{ - {x^2}}},x = 0,y = 0,x = 1
We can consider S=01ex2dxS = \int\limits_0^1 {{e^{ - {x^2}}}} dx.
x[0,1]x2x\Rightarrow x \in [0,1] \Rightarrow {x^2} \leqslant x
Multiplying both sides by 1 - 1 and taking exponents we have,
x2xex2ex\Rightarrow - {x^2} \geqslant - x \Rightarrow {e^{ - {x^2}}} \geqslant {e^{ - x}}
Therefore, S=01ex2dx01exdxS = \int\limits_0^1 {{e^{ - {x^2}}}} dx \geqslant \int\limits_0^1 {{e^{ - x}}} dx
Integrating we get, S[ex]01=e0e1=11eS \geqslant [ - {e^{ - x}}]_0^1 = {e^0} - {e^{ - 1}} = 1 - \dfrac{1}{e}
S11e\Rightarrow S \geqslant 1 - \dfrac{1}{e}
Therefore option B is also right.
To show that option C is incorrect,
e>e1e<1e\Rightarrow e > \sqrt e \Rightarrow \dfrac{1}{e} < \dfrac{1}{{\sqrt e }}
1e<1e11e<1+1e<14(1+1e)\Rightarrow - \dfrac{1}{e} < \dfrac{1}{{\sqrt e }} \Rightarrow 1 - \dfrac{1}{e} < 1 + \dfrac{1}{{\sqrt e }} < \dfrac{1}{4}(1 + \dfrac{1}{{\sqrt e }})
This implies 11e<14(1+1e)1 - \dfrac{1}{e} < \dfrac{1}{4}(1 + \dfrac{1}{{\sqrt e }})
From B, we have S11eS \geqslant 1 - \dfrac{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)S(1,0),
Clearly SS \leqslant Area of OAPQ++ Area of QBRS
PQ is drawn parallel to Y axis at the point x=12x = \dfrac{1}{{\sqrt 2 }}.
Then y=ex2=e12=1ey = {e^{ - {x^2}}} = {e^{ - \dfrac{1}{2}}} = \dfrac{1}{{\sqrt e }}.
Therefore, B is the point (12,1e)(\dfrac{1}{{\sqrt 2 }},\dfrac{1}{{\sqrt e }}).
Area of OAPQ =lenth×breadth=1×12=12 = lenth \times breadth = 1 \times \dfrac{1}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}
In QBRS, QB=1e,BR=112QB = \dfrac{1}{{\sqrt e }},BR = 1 - \dfrac{1}{{\sqrt 2 }}
Therefore, Area of QBRS =length×breadth=1e×112 = length \times breadth = \dfrac{1}{{\sqrt e }} \times 1 - \dfrac{1}{{\sqrt 2 }}
SAreaOAPQ+AreaQBRS=12+1e(112)\Rightarrow S \leqslant AreaOAPQ + AreaQBRS = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt e }}(1 - \dfrac{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.