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Mathematics Question on applications of integrals

If SS be the area of the region enclosed by y=e−x2y = e^{-x^2}, y=0,x=0y = 0,x = 0 and x=1x = 1 . Then,

A

(a)S≥1e(a) S \ge \frac{1}{e}

B

(b)S≥1−1e(b) S \ge 1 - \frac{1}{e}

C

(c)S≤14(1+1e) (c) S \le \frac{1}{4} \bigg (1 + \frac{1}{\sqrt e}\bigg)

D

(d)S≤12+1e(1−12) (d) S \le \frac{1}{\sqrt 2} + \frac{1}{\sqrt e} \bigg (1 - \frac{1}{\sqrt 2}\bigg)

Answer

(d)S≤12+1e(1−12) (d) S \le \frac{1}{\sqrt 2} + \frac{1}{\sqrt e} \bigg (1 - \frac{1}{\sqrt 2}\bigg)

Explanation

Solution

(i) Area of region f(x) bounded between x =a to x = b is
    ∫abf(x)dx\, \, \, \, \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
     ∫abf(x)dx≥∫abg(x)dx\, \, \, \, \, \int \limits_a^b f(x) dx \ge \int \limits_a^b g(x) dx
Description of Situation As the given curve y=e−x2y = e^{-x^2}
cannot be integrated, thus we have to bound this function by
using above mentioned concept.
Graph for y=e−x2y = e^{-x^2}
Since, x2≤x  when  x∈[0,1]x^2 \le x\, \, when\, \, x \in [0 , 1]
⇒−x2≥−x  or  e−x2≥e−x\Rightarrow - x^2 \ge -x\, \, or\, \, e^{-x^2} \ge e^{-x}
∴   ∫01e−x2dx≥∫01e−xdx\therefore \, \, \, \int \limits_0^1 e^{-x^2} dx \ge \int \limits_0^1 e^{-x} dx
⇒      S≥−(e−x)01=1−1e   .........(i)\Rightarrow \, \, \, \, \, \, S \ge -(e^{-x})_0^1 = 1 - \frac{1}{e} \, \, \, .........(i)
Also, ∫01e−x2dx≤\int \limits_0^1 e^{-x^2} dx \le Area of two rectangles
          ≤(1×12)+(1−12)×1e\, \, \, \, \, \, \, \, \, \, \le \bigg(1 \times \frac{1}{\sqrt2}\bigg) + \bigg(1 - \frac{1}{\sqrt2}\bigg) \times \frac{1}{\sqrt e}
        ≤12+1e(1−12)    ....(ii)\, \, \, \, \, \, \, \, \le \frac{1}{\sqrt2} + \frac{1}{\sqrt e} \bigg(1 - \frac{1}{\sqrt2}\bigg)\, \, \, \, ....(ii)
∴12+1e(1−12)≥  S  ≥1−1e   [fromEqs.(i)and(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)]