The area of the region bounded by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is

The figure shows a portion of the graph $y=2x-4x^3$. The line $y=c$ is such that the areas of the regions marked $I$ and $II$ are equal. If $a$ and $b$ are the $x$-coordinates of $A$ and $B$ respectively,then $a+b$ equals:

The area enclosed by the curve $y = 2x^2$ and the lines $x = 1$ and $y = 4$ in the first quadrant is ..... sq. units.

The area bounded by the curves $y=|x-1|+|x-2|$ and $y=3$ is equal to

The area of the region lying between the curves $y=\sqrt{4-x^2}$,$y^2=3x$ and the $Y$-axis is

For $a>0$, let the curves $C_1: y^2=a x$ and $C _2: x ^2=$ ay intersect at origin O and a point P Let the line $x = b (0 < b < a )$ intersect the chord $O P$ and the x -axis at points Q and R , respectively. If the line $x=b$ bisects the area bounded by the curves, $C _1$ and $C _2$, and the area of $\Delta OQR =\frac{1}{2}$, then ' $a$ ' satisfies the equation