Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

Using integration,find the area of the region bounded by the triangle whose vertices are $(-1, 0)$,$(1, 3)$,and $(3, 2)$.

Area of the region bounded by the curve $y=x|x|$,$X$-axis and lines $x=0$ and $x=1$ is . . . . . . sq. units.

The area of the region lying in the first quadrant and bounded by the curve $y = 4x^2$, the $Y$-axis, and the lines $y = 2$ and $y = 4$ is:

The area of the region bounded by the curve $y^2 = 4x$,the $Y$-axis,and the line $y = 3$ is . . . . . . sq. 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).