Let `F(x)=int_x^[x^2+pi/6][2cos^2t.dt]` for all `x in R` and `f:[0,1/2] -> [0,oo)` be a continuous function.For `a in [0,1/2]`, if F'(a)+2 is the area of the region bounded by x=0,y=0,y=f(x) and x=a, then f(0) is

Let F(x)=int_(x)^(x^(2)+(pi)/(6))[2cos^(2)t.dt] for all x in R and f:[0,(1)/(2)]rarr[0,oo) be a continuous function.For a in[0,(1)/(2)], if F'(a)+2 is the area of the region bounded by x=0,y=0 is the area x=a, then f(0) is

Let f : [-1, 2]to [0, oo) be a continuous function such that f(x)=f(1-x), AA x in [-1, 2] . If R_(1)=int_(-1)^(2)xf(x)dx and R_(2) are the area of the region bounded by y=f(x), x=-1, x=2 and the X-axis. Then :

Let f(x) be a continuous function such that int_(-2)^(2)f(x)dx=0 and int_(0)^(2)f(x)dx=2. The area bounded by y=f(x) , x -axis, x=-2 and x=2 is

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

Let f(x) be a continuous function defined from [0,2]rarr R and satisfying the equation int_(0)^(2)f(x)(x-f(x))dx=(2)/(3) then the value of 2f(1)