Find the area of the region bounded by $y^{2}=9x$,$x=2$,$x=4$ and the $x$-axis in the first quadrant.

The curve $y=ax^2+bx$ passes through the point $(1,2)$ and lies above the $X$-axis for $0 \leq x \leq 8$. If the area enclosed by this curve,the $X$-axis and the line $x=6$ is $108$ square units,then $2b-a=$

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

The area of the region bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and its latus rectum is . . . . . . sq. units. (where $e$ denotes the eccentricity of the ellipse).

The area inside the parabola $y^2 = 4ax$,between the lines $x = a$ and $x = 4a$ is equal to

Find the area bounded by the curves $(x-1)^{2}+y^{2}=1$ and $x^{2}+y^{2}=1$.