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

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

The area bounded by y=x+1 and y=cos x and the x - axis, is

Let f(x)=min[tan x,cot x,1sqrt(3)] for all x in(0,(pi)/(2)) 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) = minimum (tan x, cot x, 1//sqrt(3) ) AA x in [0, pi//2] . Then area bounded by y = f(x) and the x-axis is log ((a)/(b)) + (pi)/(c sqrt(d)) , then a + b - c + d = _______

Let f(x)=Min {x^(2)+1,x+1} for 0 <= x <= 2 .If A is the area bounded by y=f(x) , the x -axis in the interval [0,2] then find the integral part of A