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

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(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

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$\frac{1}{2}$

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(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

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