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

The area of the region bounded by the curve $y = \tan x$,the $X$-axis,and the line $x = \frac{\pi}{3}$ is

Area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$ is ......... $sq. \text{ unit}$.

Let $f: R \rightarrow R$ be a twice differentiable function such that $f(x + y) = f(x) f(y)$ for all $x, y \in R$. If $f^{\prime}(0) = 4a$ and $f$ satisfies $f^{\prime \prime}(x) - 3a f^{\prime}(x) - f(x) = 0$,$a > 0$,then the area of the region $R = \{(x, y) \mid 0 \leq y \leq f(ax), 0 \leq x \leq 2\}$ is:

The area of the region (in sq. units) enclosed by the curves $y=8x^3-1$,$y=0$,$x=-1$ and $x=1$ is

The area between $x=y^{2}$ and $x=4$ is divided into two equal parts by the line $x=a$. Find the value of $a$.