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  4. The area (in sq. units) bounded by [|x|]+[|y|]=2 in the first and third quadrant is (where, [.] is the greatest integer function),

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Q. The area (in sq. units) bounded by $\left[\left|x\right|\right]+\left[\left|y\right|\right]=2$ in the first and third quadrant is (where, $\left[.\right]$ is the greatest integer function),

NTA AbhyasNTA Abhyas 2020Application of Integrals

Solution:

Solution
$\left[x\right]+\left[y\right]=2$ (Let $x,y\geq 0$ )
$[x]+[y]=2$ (Let $x, y \geq 0)$
$\Rightarrow [x],[y]=2,0$ or 1,1 or 0,2
If $[x]=2$ and $[y]=0$ $\Rightarrow x \in[2,3)$ and $y \in[0,1)$ and so on Since, the curve is symmetric in the $I ^{ st }$ and $III ^{ rd }$ quadrant
Hence, the required area $=6(1 \times 1)=6$ sq. units