" Area enclosed by "f(x)=en(x-2)-(1)/(x),x=6,x=10" and "x" -axis is : "

Area enclosed by f(x)=ln(x-2)-(1)/(x),x=6,x=10 and x axis is :

Find area bounded by y = f^(-1)(x), x = 4, x = 10 and the x -axis, given that the area bounded by y = f(x) , x = 2, x = 6 and the x -axis is 30 sq. units, where f(2) = 4 and f(6) = 10 , and that f(x) is an invertible function.

The area enclosed by y = x(x - 1)(x - 2) and x-axis, is given by

The area enclosed by y = x(x - 1)(x - 2) and x-axis, is given by

If the area enclosed by g(x),x=-3,x=5 and x -axis where g(x) is the inverse f(x)=x^(3)+3x+1 is A, then [A] is

Area enclosed by the curve f(x)=((x^(2))/(4)-(8)/(x^(2)+4)) the x-axis amd ordinates x=+-3 equals

The area enclosed by y=x(x-1)(x-2) and x -axis,is given by