Estimation of Area Under Gaussian Type Curve

The area $S$ of the region bounded by $y = e^{-x^2}$, $y = 0$, $x = 0$, and $x = 1$ is given by the definite integral: $S = \int_0^1 e^{-x^2} \, dx$

Let us analyze each of the given options:

1. Analysis of Option (A): For $x \in [0, 1]$, we have: $x^2 \le 1 \implies -x^2 \ge -1$ Since the exponential function $f(t) = e^t$ is strictly increasing, it follows that: $e^{-x^2} \ge e^{-1} = \frac{1}{e}$ Integrating both sides over the interval $[0, 1]$: $S = \int_0^1 e^{-x^2} \, dx \ge \int_0^1 \frac{1}{e} \, dx = \frac{1}{e}$ Thus, $S \ge \frac{1}{e}$. Therefore, Option (A) is correct.

2. Analysis of Option (B): For $x \in [0, 1]$, we have: $x^2 \le x \implies -x^2 \ge -x$ Taking the exponential of both sides: $e^{-x^2} \ge e^{-x}$ Integrating both sides over the interval $[0, 1]$: $S = \int_0^1 e^{-x^2} \, dx \ge \int_0^1 e^{-x} \, dx = \left[ -e^{-x} \right]_0^1 = 1 - \frac{1}{e}$ Thus, $S \ge 1 - \frac{1}{e}$. Therefore, Option (B) is correct.

3. Analysis of Option (C): Numerically, we have: $1 - \frac{1}{e} \approx 1 - 0.3679 = 0.6321$ From Option (B), we know $S \ge 1 - \frac{1}{e} > 0.63$. Now consider the value in Option (C): $\frac{1}{4}\left(1 + \frac{1}{\sqrt{e}}\right) \approx \frac{1}{4}\left(1 + 0.6065\right) = \frac{1.6065}{4} \approx 0.4016$ Since $S > 0.63$, the inequality $S \le 0.4016$ is false. Therefore, Option (C) is incorrect.

4. Analysis of Option (D): We can split the integral $S$ at $x = \frac{1}{\sqrt{2}}$: $S = \int_0^{1/\sqrt{2}} e^{-x^2} \, dx + \int_{1/\sqrt{2}}^1 e^{-x^2} \, dx$

• On the interval $\left[0, \frac{1}{\sqrt{2}}\right]$, $x \ge 0 \implies x^2 \ge 0 \implies e^{-x^2} \le e^0 = 1$. Thus: $\int_0^{1/\sqrt{2}} e^{-x^2} \, dx \le \int_0^{1/\sqrt{2}} 1 \, dx = \frac{1}{\sqrt{2}}$
• On the interval $\left[\frac{1}{\sqrt{2}}, 1\right]$, $x \ge \frac{1}{\sqrt{2}} \implies x^2 \ge \frac{1}{2} \implies e^{-x^2} \le e^{-1/2} = \frac{1}{\sqrt{e}}$. Thus: $\int_{1/\sqrt{2}}^1 e^{-x^2} \, dx \le \int_{1/\sqrt{2}}^1 \frac{1}{\sqrt{e}} \, dx = \frac{1}{\sqrt{e}}\left(1 - \frac{1}{\sqrt{2}}\right)$

Summing both parts: $S \le \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{e}}\left(1 - \frac{1}{\sqrt{2}}\right)$ Therefore, Option (D) is correct.

Hence, the correct options are A, B, and D.