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

The area lying between the curves $y^{2}=4x$ and $y=2x$ is

The area of the region bounded by the parabola $y^2 = 8x$ and its latus rectum is . . . . . . sq. units.

The area bounded by $y=x^{3}$,$y=8$,and $x=0$ is

The area of the region $\{(x, y): 0 \leq x \leq \frac{9}{4}, 0 \leq y \leq 1, x \geq 3y, x+y \geq 2\}$ is

Let $n \geq 2$ be a natural number and $f:[0,1] \rightarrow \mathbb{R}$ be the function defined by
$f(x)= \begin{cases} n(1-2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n(2nx-1) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1-nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n-1}(nx-1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}$
If $n$ is such that the area of the region bounded by the curves $x=0, x=1, y=0$ and $y=f(x)$ is $4$,then the maximum value of the function $f$ is