" Let "f(x)=min{(x+1),sqrt((1-x))}" ,then area bounded by "f(x)" and "x" -axis is: "

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

f(x) = min { x+1 , sqrt((1-x))} the area of the region bounded by the curve f(x) and X - axis is ….Sq. unit

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

Find the area bounded by the curve f(x) =4-|x| and the x axis

Let f(x)=min(x+1,x,1-x). Then f(2) equals

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