The area of the region in the first quadrant inside the circle $x^2+y^2=8$ and outside the parabola $y^2=2x$ is equal to:

The area of the region (in square units) bounded by the curves $y=x^3$,$y=x$ and $-1 \leq x \leq 1$ is:

Let $A_1$ be the bounded area enclosed by the curves $y=x^2+2$,$x+y=8$ and the y-axis that lies in the first quadrant. Let $A_2$ be the bounded area enclosed by the curves $y=x^2+2$,$y^2=x$,$x=2$,and the y-axis that lies in the first quadrant. Then $A_1-A_2$ is equal to

The area of the region bounded by the curves $y^{2}=8x$ and $y=x$ is

The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ into two parts. The area of the smaller part is equal to:

If the area (in $sq. units$) of the region $\{(x,y): y^2 \le 4x, x + y \le 1, x \ge 0, y \ge 0\}$ is $a\sqrt{2} + b$,then $a - b$ is equal to