If the area enclosed by curve `y=f(x)a n dy=x^2+2` between the abscissa `x=2a n dx=alpha,alpha>2,` is `(alpha^3-4alpha^2+8)s qdot` unit. It is known that curve `y=f(x)` lies below the parabola `y=x^2+2.`

The area enclosed between the curves y^(2)=x and y=|x| is

find the area enclosed between the curves y=x^(2)-5x and y=4-2x

The area enclosed between the curves y=x and y=2x-x^(2) (in square units), is

Area enclosed between the curve y=x^(2) and the line y = x is

Area enclosed between the curve y^(2)=x and the line y = x is

The area enclosed between the curve y^(2)(2a-x)=x^(3) and the line x=2 above the x-axis is

Find the value of alpha for which the point (alpha -1, alpha) lies inside the parabola y^(2) = 4x .

If f(x) is a non - negative function such that the area bounded by y=f(x), x - axis and the lines x = 0 and x=alpha is 4alpha sin alpha+2 sq. Units ( AA alpha in [0, pi] ), then the value of f((pi)/(2)) is equal to