Find the area bounded by the curves $\{(x, y): y \geq x^{2} \text{ and } y=|x|\}$.

The area bounded by the circle $x^2 + y^2 = 4$,the line $x = \sqrt{3}y$,and the $x$-axis lying in the first quadrant is:

The area enclosed (in square units) by the curve $y=x^4-x^2$, the $x$-axis and the vertical lines passing through the two minimum points of the curve is

The area of the region bounded by the curve $y = \cos x$ between $x = 0$ and $x = \frac{3\pi}{2}$ is . . . . . . square units.

Let $A$ be the area bounded by the curve $y = \cos^{-1}\sqrt{1 - x^2}$,the tangent to the curve $y = \sin^{-1}x$ at $x = 0$,and the line $x = 1$. Then the value of $2(\{A\} + \text{sgn}(A))$ is (where $\{.\}$ is the fractional part function and $\text{sgn}(x)$ is the signum function).

If a curve $y=a \sqrt{x}+b x$ passes through the point $(1,2)$ and the area bounded by this curve,line $x=4$ and the $X$-axis is $8$ sq.units,then the value of $a-b$ is