f(x)=sqrt(1-x^(2)),0<=x<=1

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

The function f(x)=(sqrt(1+x^(2))-sqrt(1-x^(2))/(x^(2))"for "x ne 0, f(0) =1 at x=0 is

Can Lagrange's Theorem be applied to the function f(x) = sqrt(1-x^2) in the interval [0,1]

Let f(x) be defined in [-2,2] by f(x)={max(sqrt(4)-x^(2)),sqrt(1+x^(2))),-2<=x<=0;min(sqrt(4-x^(2)),sqrt(1+x^(2)),0

If f(x) + f(y) = f(xy-sqrt(1-x^(2))sqrt(1-y^(2))), f(0) = (pi)/(2) and f is differentiable in (-1, 1). Then |lim_(x rarr 0)(2f(x)-pi)/(x)|=

If f(x) + f(y) = f(xy-sqrt(1-x^(2))sqrt(1-y^(2))), f(0) = (pi)/(2) and f is differentiable in (-1, 1). Then |lim_(x rarr 0)(2f(x)-pi)/(x)|=

If f is continuous at x=0, where f(x)=x^(2)+alpha,x>=0f(x)=2sqrt(x^(2)+1)+beta,x<0. Find alpha and beta given that f((1)/(2))=2

If int1/((1+x^(2))sqrt(1-x^(2)))dx=F(x) " and " F(1)=0 , then for x gt 0, F(x)=