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(B) The area $S$ is given by the integral $S = \int_0^1 e^{-x^2} dx$.$1$. For $x \in [0, 1]$,we have $0 \leq x^2 \leq x \leq 1$. Thus,$-x^2 \geq -x$,w

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$(A, B, D)$

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(B) The area $S$ is given by the integral $S = \int_0^1 e^{-x^2} dx$.$1$. For $x \in [0, 1]$,we have $0 \leq x^2 \leq x \leq 1$. Thus,$-x^2 \geq -x$,which implies $e^{-x^2} \geq e^{-x}$.Integrating both sides from $0$ to $1$:$S = \int_0^1 e^{-x^2} dx \geq \int_0^1 e^{-x} dx = [-e^{-x}]_0^1 = 1 - \frac{1}{e}$.Since $1 - \frac{1}{e} \approx 1 - 0.367 = 0.633$ and $\frac{1}{e} \approx 0.367$,we have $S \geq 1 - \frac{1}{e} > \frac{1}{e}$. Thus,$(A)$ and $(B)$ are correct.$2$. Using the upper Riemann sum with two rectangles of width $\frac{1}{\sqrt{2}}$ and $1-\frac{1}{\sqrt{2}}$:The function $f(x) = e^{-x^2}$ is decreasing on $[0, 1]$.$S = \int_0^{1/\sqrt{2}} e^{-x^2} dx + \int_{1/\sqrt{2}}^1 e^{-x^2} dx$.Using the maximum value of $f(x)$ on each sub-interval:$S \leq \left(\frac{1}{\sqrt{2}}\right) f(0) + \left(1 - \frac{1}{\sqrt{2}}\right) f\left(\frac{1}{\sqrt{2}}\right)$$S \leq \frac{1}{\sqrt{2}}(1) + \left(1 - \frac{1}{\sqrt{2}}\right) e^{-(1/\sqrt{2})^2} = \frac{1}{\sqrt{2}} + \left(1 - \frac{1}{\sqrt{2}}\right) \frac{1}{\sqrt{e}}$.Thus,$(D)$ is correct.

If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

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If $S$ be the area of the region enclosed by $y=e^{-x^2}, y=0, x=0$,and $x=1$. Then$(A) S \geq \frac{1}{e}$$(B) S \geq 1-\frac{1}{e}$$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$