Area bounded by `f(x)=1+sqrt[1-x^2]`, ` x in [0,1]` and `f^-1(x)` with coordinate axes is

Area bounded by f(x)=(x^(2)-1)/(x^(2)+1) and the line y = 1 is

Area bounded by f(x)=(x^(2)-1)/(x^(2)+1) and the line y = 1 is

Area bounded by f(x)=(x^(2)-1)/(x^(2)+1) and the line y = 1 is

If f(x) = sqrt(1-sqrt1-x^2 , then at x = 0

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

The area bounded by the curve y=f(x) (where f(x)>=0) ,the co-ordinate axes & the line x=x_(1) is given by x_(1).e^(x_(1)). Therefore f(x) equals

The area bounded by f(x)=min_( min )(|x|,|x-1|,|x+1|) in [-1,1] and X-axis,is

f(x)=sqrt(1-sqrt(1-x^2) then at x=0 ,value of f(x) is