f(x) = minimum of `{|x – 1|,|x|,|x +1|}` and `x= +-1` the area of f(x) is in sq.units

f(x)= minimum of {x-[x],-x-[-x]} and x=0 and x=4 then the area of f(x) in sq.units

f(x)=min {x-[x],-x-[-x]} and x=0 and x=4 then the area of f(x) in sq. units

Let f(x)=min{|x+2|,|x|,|x-2|} then the value of the integral int_-2^2 f(x)dx is (i) 4 sq.unit (ii)10 sq.unit (iii)2 sq.unit (iv) 7 sq.unit

If A ={(x,y} ,x^2 +y^2 le 1 and y^2 le 1 -x} then the area of A is …..Sq. units .

Let f(x)= minimum (x+1,sqrt(1-x)) for all xle1 . Then the area bounded by y=f(x) and the x-axis is a) (7)/(3) sq. units b) 1/6 sq. units c) 11/6 sq. units d) 7/6 sq. units

Let f(x) = minimum (x+1, sqrt(1-x))" for all "x le 1. Then the area bounded by y=f(x) and the x-axis is

Let f(x) = minimum (x+1, sqrt(1-x))" for all "x le 1. Then the area bounded by y=f(x) and the x-"axis" is

Let f(x) = minimum (x+1, sqrt(1-x))" for all "x le 1. Then the area bounded by y=f(x) and the x-"axis" is

Let f(x)= minimum (x+1,sqrt(1-x) for all x<=1 then the area bounded by y=f(x) and the x -axis,is