The area (in square units) bounded by the curves $y = \sqrt{x}$,the line $2y - x + 3 = 0$,and the $X$-axis,lying in the first quadrant,is:

The area of the region enclosed by the parabola $(y-2)^2=x-1$,the line $x-2y+4=0$,and the positive coordinate axes is

Let $P_{1}: y = 4x^{2}$ and $P_{2}: y = x^{2} + 27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y = \alpha x, \alpha > 0$ and $P_{1}$,then $\alpha$ is equal to:

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

The area enclosed by $y^{2}=8x$ and $y=\sqrt{2}x$ that lies outside the triangle formed by $y=\sqrt{2}x$,$x=1$,and $y=2\sqrt{2}$ is equal to

Find the area enclosed by the parabola $4y = 3x^{2}$ and the line $2y = 3x + 12$.