Let the area of the bounded region $\{(x, y): 0 \leq 9x \leq y^2, y \geq 3x-6\}$ be $A$. Then $6A$ is equal to . . . . . .

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 area bounded by the curve $y = x^2 + 2$,the $x$-axis,and the lines $x = 1$ and $x = 2$ is:

Find the area of the region bounded by the line $y=3x+2$,the $x$-axis,and the ordinates $x=-1$ and $x=1$.

The area bounded by the curve $y = 2x^2$,the $X$-axis,and the line $x = 1$ is . . . . . . sq. units.

The area bounded by the parabola ${y^2} = 4ax$ and its latus rectum is: