If S be the area of the region enclosed by y= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$(A, B, D)$

Vedclass · Answer

(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
$(A) S \geq \frac{1}{e}$
$(B) S \geq 1-\frac{1}{e}$
$(C) S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
$(D) S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

Similar Questions

The area of the region enclosed by the curve $f(x) = \max \{\sin x, \cos x\}$,$-\pi \leq x \leq \pi$ and the $x$-axis is

The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$, $y=\cos x$ and the $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$ is

The area of the region bounded by the curve $y=x|x|$,lines $x=-1$ and $x=1$ is . . . . . . .

The area of the region bounded by the circle $x^2+y^2=16$ and the lines $x=0$ and $x=4$ in the first quadrant is . . . . . . sq. units. (in $\pi$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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)$