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

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

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

Consider f(x)= minimum (x+2, sqrt(4-x)), AA x le 4 . If the area bounded by y=f(x) and the x - axis is (22)/(k) square units, then the value of k 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)=min(x+1,sqrt(1-x))AA x le 1 . Then, the area (in sq. units( bounded by y=f(x), y=0 and x=0 from y=0 to x=1 is equal to

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

Let f(x)={{:(2^([x]),, x lt0),(2^([-x]),,x ge0):} , then the area bounded by y=f(x) and y=0 is