The area of the region bounded by $y=-\sqrt{16-x^{2}}$ and the $X$-axis is

Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$. (in $\pi$)

The area bounded by the circle $x^{2}+y^{2}=16$ and the lines $x=0$ and $x=2$ 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

The graphs of sine and cosine functions intersect each other at a number of points,and between two consecutive points of intersection,the two graphs enclose the same area $A$. Then $A^{4}$ is equal to ............

The area of the region bounded by the curves $y = |x - 4|$, $x = 3$, $x = 5$, and the $X$-axis is