Q.
The area (in sq. units) bounded by [∣x∣]+[∣y∣]=2 in the first and third quadrant is (where, [.] is the greatest integer function),
2141
190
NTA AbhyasNTA Abhyas 2020Application of Integrals
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Solution:
[x]+[y]=2 (Let x,y≥0 ) [x]+[y]=2 (Let x,y≥0) ⇒[x],[y]=2,0 or 1,1 or 0,2
If [x]=2 and [y]=0⇒x∈[2,3) and y∈[0,1) and so on Since, the curve is symmetric in the Ist and IIIrd quadrant
Hence, the required area =6(1×1)=6 sq. units