The area of the region bounded by the curves $y=|x-1|$ and $y=3-|x|$ is

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Aicharya · Accepted Answer

$A=\int_{-1}^0\{(3+x)-(-x+1)\} d x+\int_0^1\{(3-x)-(-x+1)\} d x+\int_1^2\{(3-x)-(-x-1)\} d x$ $=\int_{-1}^0(2+2 x) d x+\int_0^1 2 d x+\int_1^2(4-2 x) d x$ $=\left[2 x-x^2\right]_{-1}^0+[2 x]_0^1+\left[4 x-x^2\right]_1^2$ $=0-(-2+1)+(2-0)+(8-4)-(4-1)$ $=1+2+4-3=4$ sq. units

The area of the region bounded by the curves $y=|x-1|$ and $y=3-|x|$ is

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