The area of the region bounded by y=|x| and y | vedclass

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

Question

(B) The given curves are $y=|x|$ and $y=1-|x|$.To find the points of intersection, set $|x| = 1-|x|$, which gives $2|x| = 1$, so $|x| = \frac{1}{2}$.T

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(B) The given curves are $y=|x|$ and $y=1-|x|$.To find the points of intersection, set $|x| = 1-|x|$, which gives $2|x| = 1$, so $|x| = \frac{1}{2}$.This implies $x = \frac{1}{2}$ or $x = -\frac{1}{2}$.When $x = \frac{1}{2}$, $y = \frac{1}{2}$. When $x = -\frac{1}{2}$, $y = \frac{1}{2}$.The intersection points are $A(\frac{1}{2}, \frac{1}{2})$ and $C(-\frac{1}{2}, \frac{1}{2})$.The curves also intersect the $y$-axis at $O(0,0)$ and $B(0,1)$.The region is a square with vertices $O(0,0)$, $A(\frac{1}{2}, \frac{1}{2})$, $B(0,1)$, and $C(-\frac{1}{2}, \frac{1}{2})$.The side length of the square is the distance $OA = \sqrt{(\frac{1}{2}-0)^2 + (\frac{1}{2}-0)^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.The area of the square is $(	ext{side})^2 = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2} 	ext{ sq unit}$.

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

Similar Questions

The volume of a spherical cap of height $h$ cut off from a sphere of radius $a$ is equal to

Find the area under the given curve $y=x^{2}$ bounded by the lines $x=1$,$x=2$ and the $x$-axis.

The area of the region bounded by the parabola $y=x^2$ and the line $y=4$ is . . . . . . sq. units.

If a straight line $y - x = 2$ divides the region $x^2 + y^2 \le 4$ into two parts,then the ratio of the area of the smaller part to the area of the greater part is

The area of the curve $xy^2 = a^2(a - x)$ bounded by the $y$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module