The area bounded by y-1=-|x| and y+1=|x| is | vedclass

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

Question

(C) The given equations are $y = 1 - |x|$ and $y = |x| - 1$.These equations represent a square with vertices at $A(0, 1)$,$C(1, 0)$,$B(0, -1)$,and $D(

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equations are $y = 1 - |x|$ and $y = |x| - 1$.These equations represent a square with vertices at $A(0, 1)$,$C(1, 0)$,$B(0, -1)$,and $D(-1, 0)$.The area of the region bounded by these curves is the area of the square $ACBD$.The area of a square with vertices $(0, 1)$,$(1, 0)$,$(0, -1)$,and $(-1, 0)$ can be calculated by dividing it into four congruent right-angled triangles,each with base $1$ and height $1$.Area of one triangle (e.g.,$	riangle AOC$) $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$.Total area $= 4 	imes 	ext{Area of } 	riangle AOC = 4 	imes \frac{1}{2} = 2$.Thus,the area of the required bounded region is $2$ square units.

The area bounded by $y-1=-|x|$ and $y+1=|x|$ is

Similar Questions

The area of the region $A = \{(x, y) : |\cos x - \sin x| \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2}\}$ is:

The area of the region $\{(x, y) : 0 \le y \le 6 - x, y^2 \ge 4x - 3, x \ge 0\}$ is:

The area of the region enclosed by the parabola $(y-2)^2=x-1$,the line $x-2y+4=0$,and the positive coordinate axes is

The area bounded by the curves $y = \cos x$ and $y = \sin x$ between the ordinates $x = 0$ and $x = \frac{3\pi}{2}$ is:

Let $f :[-3,1] \rightarrow R$ be given as
$f(x)=\begin{cases} \min \{(x+6), x^{2}\}, & -3 \leq x \leq 0 \\ \max \{\sqrt{x}, x^{2}\}, & 0 \leq x \leq 1 \end{cases}$
If the area bounded by $y = f(x)$ and the $x$-axis is $A$,then the value of $6A$ is equal to ....... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module