Find the area of the parabola $y^{2}=4ax$ bounded by its latus rectum.

The area of the region bounded by the lines $2x - y + 1 = 0$, $y = -1$, $y = 3$ and the y-axis is

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

The area of the region bounded by the curve $y = \cos x$,$x = 0$,and $x = 3\pi$ is . . . . . . sq. units.

If the line $x=\alpha$ divides the area of region $R=\{(x, y) \in \mathbb{R}^2: x^3 \leq y \leq x, 0 \leq x \leq 1\}$ into two equal parts,then which of the following is true?
$[A] \ 0 < \alpha \leq \frac{1}{2}$
$[B] \ \frac{1}{2} < \alpha < 1$
$[C] \ 2 \alpha^4 - 4 \alpha^2 + 1 = 0$
$[D] \ \alpha^4 + 4 \alpha^2 - 1 = 0$

Find the area of the region bounded by the curve $y^{2}=x$,the lines $x=1$,$x=4$,and the $x$-axis in the first quadrant.