The area of the region bounded by the curve $y=x$,the lines $x=1$ and $x=10$ using integration is . . . . . . sq. units.

Find the area enclosed by the circle $x^{2}+y^{2}=a^{2}$.

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 . . . . . .

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 line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x$, $y=\cos x$ and the $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1 : A_2$ equals (in $: 1$)

The area under the curve $y = x^2 - 4x$ within the $x$-axis and the line $x = 2$ is: